(navigation image)
Home American Libraries | Canadian Libraries | Universal Library | Community Texts | Project Gutenberg | Children's Library | Biodiversity Heritage Library | Additional Collections
Search: Advanced Search
Anonymous User (login or join us)
Upload
See other formats

Full text of "Elementary Differential Geometry"

BARRETT 

O'NEILL 

M M 

* L 

PC H 



BARRETT O'NEILL 



ELEME NTARY 
DIFFERE NTIAL 

GEOMETRY 



>v*u 



Ui 



111 



h> 



ft(p) 




(if) 



* 



Af 



— Ar 







Elementary 
Differential Geometry 



Elementary 
Differential Geometry 

Barrett O'Neill 



DEPARTMENT OF MATHEMATICS 
UNIVERSITY OF CALIFORNIA 
LOS ANGELES, CALIFORNIA 




ACADEMIC PRESS New York San Francisco London 
A Subsidiary of Harcourt Brace Jovanovich, Publishers 



Copyright © 1966, by Academic Press, Inc. 

all rights reserved. 

no part of this publication may be reproduced or 

transmitted in any form or by any means, electronic 

or mechanical, including photocopy, recording, or any 

information storage and retrieval system, without 

permission in writing from the eublisher. 



ACADEMIC PRESS, INC. 

Ill Fifth Avenue, New York, New York 10003 



United Kingdom Edition published by 
ACADEMIC PRESS, INC. (LONDON) LTD. 
24/28 Oval Road, London NW1 



Library of Congress Catalog Card Number: 66-14468 



PRINTED IN THE UNITED STATES OF AMERICA 



Preface 



This book is an elementary account of the geometry of curves and surfaces. 
It is written for students who have completed standard first courses in 
calculus and linear algebra, and its aim is to introduce some of the main 
ideas of differential geometry. 

The traditional undergraduate course in differential geometry has changed 
very little in the last few decades. By contrast, geometry has been ad- 
vancing very rapidly at the research level, and there is general agreement 
that the undergraduate course needs to be brought up to date. I have tried 
to think through the classical material, to prune and augment it, and to 
write down the results in a reasonably clean and modern mathematical style- 
However, I have used a new idea only if it really pays its way by simplify- 
ing and clarifying the exposition. 

Chapter I establishes the language of the book — a language compounded of 
familiar parts of calculus and linear algebra. Chapter II describes the 
method of "moving frames," which is introduced, as in elementary calculus, 
to study curves in space. In Chapter III we investigate the rigid motions 
of space, in terms of which congruence of curves (or surfaces) in space is 
defined in the same fashion as congruence of triangles in the plane. 

Chapter IV requires special comment. The main weakness of classical 
differential geometry was its lack of any adequate definition of surface. 
In this chapter we decide just what a surface is, and show that each surface 
has a differential and integral calculus of its own, strictly comparable with 
the familiar calculus of the plane. This exposition provides an introduction 
to the notion of differ entiable manifold, which has become indispensable to 
those branches of mathematics and its applications based on the calculus. 

The next two chapters are devoted to the geometry of surfaces in 3-space. 
Chapter V stresses intuitive and computational aspects to give geometrical 
meaning to the theory presented in Chapter VI. In the final chapter, al- 
though our methods are unchanged, there is a radical shift of viewpoint. 
Roughly speaking, we study the geometry of a surface as seen by its inhabi- 



vi PREFACE 

tants, with no assumption that the surface is to be found in ordinary three- 
dimensional space. 

No branch of mathematics makes a more direct appeal to the intuition 
than geometry. I have sought to emphasize this by a large number of illus- 
trations, which form an integral part of the text. A set of exercises appears 
at the end of each section; these range from routine tests of comprehension 
to more seriously challenging problems. 

In teaching from preliminary versions of this book, I have usually cov- 
ered the background material in Chapter I rather rapidly, and have not 
devoted any classroom time to Chapter III (hence also Section 8 of Chapter 
VI). A course in the geometry of curves and surfaces in space might consist 
of: Chapter II, Chapter IV (omit Sections 6 and 8), Chapter V, and Chapter 
VI (omit Sections 6 and 7). This is essentially the content of the tradi- 
tional undergraduate course in differential geometry, with clarification of 
the concepts of surface and mapping of surfaces. 

The omitted sections in the list above are used only in Chapter VII. This 
final chapter, an extensive account of two-dimensional Riemannian geom- 
etry, is in a sense the goal of the book. Rather than shift the discourse to 
higher dimensions, I have preferred to retain dimension 2, so that this more 
sophisticated view of geometry will develop directly from the special case 
of surfaces in 3-space. Chapter VII is long, and on a first reading Theorem 
5.9 and Sections 6 and 7 may well be omitted. Serious use of differential 
equations theory has been largely avoided in the early chapters; however, 
some acquaintance with the fundamentals of the subject will be helpful 
in Chapter VII. 

Los Angeles, California B. O'N. 



Contents 



Preface 



Introduction l 

Chapter I. Calculus on Euclidean Space 

1. Euclidean Space 3 

2. Tangent Vectors 6 

3. Directional Derivatives 11 

4. Curves in E 3 15 

5. 1-Forms 22 

6. Differential Forms 26 

7. Mappings 32 

8. Summary 41 

Chapter II. Frame Fields 

1. Dot Product 42 

2. Curves 51 

3. The Frenet Formulas 56 

4. Arbitrary-Speed Curves 66 

5. Covariant Derivatives 77 

6. Frame Fields 81 

7. Connection Forms 85 

8. The Structural Equations 91 

9. Summary 96 

Chapter III. Euclidean Geometry 

1. Isometries of E 3 98 

2. The Derivative Map of an Isometry 104 

3. Orientation 107 

4. Euclidean Geometry 112 

5. Congruence of Curves 116 

6. Summary 123 



vin CONTENTS 

Chapter IV. Calculus on a Surface 

1. Surfaces in E 3 124 

2. Patch Computations 133 

3. Differentiable Functions and Tangent Vectors 143 

4. Differential Forms on a Surface 152 

5. Mappings of Surfaces 158 

6. Integration of Forms 167 

7. Topological Properties of Surfaces 176 

8. Manifolds 182 

9. Summary 187 

Chapter V. Shape Operators 

1. The Shape Operator of M C E 3 189 

2. Normal Curvature 195 

3. Gaussian Curvature 203 

4. Computational Techniques 210 

5. Special Curves in a Surface 223 

6. Surfaces of Revolution 234 

7. Summary 244 

Chapter VI. Geometry of Surfaces in E 3 

1. The Fundamental Equations 245 

2. Form Computations 251 

3. Some Global Theorems 256 

4. Isometries and Local Isometries 263 

5. Intrinsic Geometry of Surfaces in E 3 271 

6. Orthogonal Coordinates 276 

7. Integration and Orientation 280 

8. Congruence of Surfaces 297 

9. Summary 303 

Chapter VII. Riemannian Geometry 

1. Geometric Surfaces 304 

2. Gaussian Curvature 310 

3. Covariant Derivative 318 

4. Geodesies 326 

5. Length-Minimizing Properties of Geodesies 339 

6. Curvature and Conjugate Points 352 

7. Mappings that Preserve Inner Products 362 

8. The Gauss-Bonnet Theorem 372 

9. Summary 389 

Bibliography 391 

Answers to Odd-Numbered Exercises 393 

Index 405 



Elementary 
Differential Geometry 



Introduction 



This book presupposes a reasonable knowledge of elementary calculus and 
linear algebra. It is a working knowledge of the fundamentals that is 
actually required. The reader will, for example, frequently be called upon 
to use the chain rule for differentiation, but its proof need not concern us. 

Calculus deals mostly with real-valued functions of one or more varia- 
bles, linear algebra with functions (linear transformations) from one 
vector space to another. We shall need functions of these and other types, 
so we give here general definitions which cover all types. 

A set S is a collection of objects which are called the elements of S. A 
set A is a subset of S provided each element of A is also an element of S. 

A function f from a set D to a set R is a rule that assigns to each element 
x of D a, unique element f(x) of R. The element f(x) is called the value 
of / at x. The set D is called the domain of /; the set R is often called the 
range of /. If we wish to emphasize the domain and range of a function /, 
the notation f:D-*R is used. Note that the function is denoted by a single 
letter, say /, while fix) is merely a value of /. 

Many different terms are used for functions — mappings, transformations, 
correspondences, operators, and so on. A function can be described in 
various ways, the simplest case being an explicit formula such as 

f(x) = 3x 2 + 1, 

which we may also write as x — > Zx + 1 . 

If both fi and f% are functions from D to R, then fr = / 2 means that 
fa( x ) = f 2 ( x ) for all x in D. This is not a definition, but a logical consequence 
of the definition of function. JJ 

Let f: D —> R and g: E — > S be functions. In general, the image oiyf is 
the subset of R consisting of all elements of the form f(x); it is usually 
denoted by /(D). Now if this image also happens to be a subset of the 
domain E of g, it is possible to combine these two functions to obtain the 
composite function g(J): D -*• S. By definition, g(f) is the function whose 
value on each element x of D is the element g(f(x)) of S. 



INTRODUCTION 



3 



If /: D — > R is a function and A is a subset of D, then the restriction of 
/ to A is the function/ | A: A — ► R denned by the same rule as/, but applied 
only to elements of A. This seems a rather minor change, but the function 
/ | A may have properties quite different from / itself. 

Here are two vital properties which a function may possess. A function 
f:D^>R is one-to-one, provided that if x and y are any elements of D such 
that x 9* y, then f(x) 9* f(y). A function f.D^R is onto (or carries D 
onto R) provided that for every element y of R there is at least one element 
x of D such that /Or) = y. In short, the image of^fs the entire set R. For 
example, consider the following functions, each of which has the real 
numbers as both domain and range : 

(1 ) The function x — ► x 3 is both one-to-one and onto. 

(2) The exponential function x — > e x is one-to-one, but not onto. 

(3) The function x — » x z + x 2 is onto, but not one-to-one. 

(4) The sine function x — > sin x is neither one-to-one nor onto. 

If a function /: D — > R is both one-to-one and onto, then for each ele- 
ment y of R there is one and only one element x such that f(x) = y.By 
defining/ -1 (y) = x for all x and y so related, we obtain a function/ -1 : R-+D 
called the inverse of/. Note that the function/ -1 is also one-to-one and onto, 
and that its inverse function is the original function /. 

Here is a short list of the main notations used throughout the book, in 
order of their appearance in Chapter I : 

P> <I> points (Sec. 1 ) 

/» 9> real-valued functions (Sec. 1 ) 

v » w » tangent vectors (Sec. 2) 

V>W, vector fields (Sec. 2) 

«> Pi curves (Sec. 4) 

<k $> differential forms (Sec. 5) 

F > G > mappings (Sec. 7) 

In Chapter I we define these concepts for Euclidean 3 -space. (Extension 
to arbitrary dimensions is virtually automatic.) In Chapter IV we show 
how these concepts can be adapted to a surface. 

A few references are given to the brief bibliography at the end of the 
book; these are indicated by numbers in square brackets. 



CHAPTER 



I 



Calculus on Euclidean Space 



As mentioned in the Preface, the purpose of this initial chapter is to 
establish the mathematical language used throughout the book. Much 
of what we do is simply a review of that part of elementary calculus dealing 
with differentiation of functions of three variables, and with curves in space. 
Our definitions have been formulated so that they will apply smoothly 
to the later study of surfaces. 



1 Euclidean Space 

Three-dimensional space is often used in mathematics without being 
formally defined. It is said to be the space of ordinary experience. Looking 
at the corner of a room, one can picture the familiar process by which 
rectangular coordinate axes are introduced and three numbers are meas- 
ured to describe the position of each point. A precise definition which 
realizes this intuitive picture may be obtained by this device: instead of 
saying that three numbers describe the position of a point, we define them 
to be a point. 

1.1 Definition Euclidean S-space E 3 is the set of all ordered triples of 
real numbers. Such a triple p = (pi, p 2 , Pz) is called a point of E 3 . 

In linear algebra, it is shown that E 3 is, in a natural way, a vector space 
over the real numbers. In fact, if p = (p\,P2,Pz) and q = {q\,qi,qz) 
are points of E 3 , their sum is the point 

P + q = (pi + gi, pt + g2, ps + gs). 

The scalar product of a point p = (p u p 2 , Pz) by a number a is the point 

ap = (api, ap 2 , ap z ). 



* CALCULUS ON EUCLIDEAN SPACE [Chap. I 

It is easy to check that these two operations satisfy the axioms for a vector 
space. The point = (0, 0, 0) is called the origin of E 3 . 

Differential calculus deals with another aspect of E 3 starting with the 
notion of differentiable real-valued functions on E 3 . We recall some funda- 
mentals. 

1 .2 Definition Let x, y, and z be the real-valued functions on E 3 such 
that for each point p = (pi,p 2 ,ps) 

z(p) = pi, y(p) = Pi, z(p) = p s . 

These functions x, y, z are called the natural coordinate functions of E 3 . 
We shall also use index notation for these functions, writing 

xi = x, x 2 = y, x z = z. 

Thus the value of the function Xi on a point p is the number p { , and so 
we have tlwydentity p = (pi,p 2 ,p 3 ) = (:ci(p), x 2 (p), x»(p)) for each 
point p of E . Elementary calculus does not always make a sharp distinc- 
tion between the numbers p u p 2 , p 3 and the functions x 1} x 2 , x 3 . Indeed the 
analogous distinction on the real line may seem pedantic, but for higher- 
dimensional spaces such as E 3 , its absence leads to serious ambiguities. 
(Essentially the same distinction is being made when we denote a function 
on E 3 by a single letter/, reserving /(p) for its value at the point p.) 

We assume that the reader is familiar with partial differentiation and 
its basic properties, in particular the chain rule for differentiation of a 
composite function. We shall work mostly with first-order partial deriva- 
tives df/dx, df/dy, df/dz and second-order partial derivatives d 2 f/dx 2 , 
d f/dx dy, • ■ • . In a few situations, third-, and even fourth-, order deriva- 
tives may occur, but to avoid worrying about exactly how many derivatives 
we can take in any given context, we establish the following definition. 

1.3 Definition A real-valued function / on E 3 is differentiable (or, in- 
finitely differentiable, or of class C°°) provided all partial derivatives of 
/, of all orders, exist and are continuous. 

Differentiable real-valued functions / and g may be added and multi- 
plied in a familiar way to yield functions that are again differentiable and 
real-valued. We simply add and multiply their values at each point— the 
formulas read 

(/+<7)(p) =/(p) + <7(p), (fg)(p) =/(pMp). 

The phrase "differentiable real-valued function" is unpleasantly long. 
Hence we make the convention that unless the context indicates otherwise, 
"function" shall mean "real-valued function," and (unless the issue is 
explicitly raised) the functions we deal with will be assumed to be dif- 



Sec. 1] EUCLIDEAN SPACE 5 

ferentiable. We do not intend to overwork this convention; for the sake 
of emphasis the words "differentiable" and "real-valued" will still appear 
fairly frequently. 

Differentiation is always a local operation : To compute the value of the 
function df/dx at a point p of E 3 , it is sufficient to know the values of / at 
all points q of E 3 which are sufficiently near p. Thus Definition 1.3 is 
unduely restrictive; the domain of /need not be the whole of E , but need 
only be an open set of E 3 . By an open set of E we mean a subset of E 
such that if a point p is in 0, then so is every other point of E that is 
sufficiently near p. (A more precise definition is given in Chap. II.) For 
example, the set of all points p = (pi, pi, pz) in E 3 such that p x > is an 
open set, and the function yz log x defined on this set is certainly differ- 
entiable, even though its domain is not the whole of E . Generally speaking, 
the results in this chapter remain valid if E 3 is replaced by an arbitrary 
open set of E . 

We are dealing with three-dimensional Euclidean space for no better 
reason than that this is the dimension we use most often in later work. It 
would be just as easy to work with Euclidean n-space E", for which the 
points are n-tuples p = (pi, • • • , p n ), and which has n natural coordinate 
functions x x , • • • , x n . All the results in this chapter are valid for Euclidean 
spaces of arbitrary dimensions, although we shall rarely take advantage 
of this except in the case of the Euclidean plane E 2 . In particular, the re- 
sults are valid for the real line E 1 = R. Many of the concepts introduced 
are specifically designed to deal with higher dimensions, however, and are 
thus apt to be overelaborate when reduced to dimension 1. 



EXERCISES 

1. Let / = x 2 y and g = y sin z be functions on E 3 . Express the following 
functions in terms of x, y, z: 

<•>"■• w %< + %'• 

b_ 

by dz N ~' by 



(C)?M. (d)i-(sin/) 



2. Find the value of the function / = x 2 y — y 2 z at each point : 
(a) (1,1,1). -' (b) (3, -1,|). 
(c) (a, 1,1 -a). (d) (t,t 2 ,f). 



6 CALCULUS ON EUCLIDEAN SPACE [Chap. I 

3. Express df/dx in terms of x, y, and z if 

(a) / = x sin (xy) + y cos (xz). 

(b) f = sing, g = e\ ft = x 2 + ?/ 2 + z 2 . 

4. If 0i, 02, ^3 and h are real-valued functions on E 3 , then 

/ = h(gi,g 2 ,g 3 ) 
is the function such that 

/(p) = A(0i(p)t 02 (p), 03 (p)) for all p.f 
Express d//dx in terms of x, y, and z, if h = x 2 — yz and 
(a) / = A(s + y, i/ 2 , x + z). 
(b) /=/*(**, e* + *,e*). 

(c) / = h(x, —x, x). 

2 Tangent Vectors 

Intuitively, a vector in E is an oriented line segment, or "arrow." Vectors 
are used widely in physics and engineering to describe forces, velocities, 
angular momenta, and many other concepts. To obtain a definition that 
is both practical and precise, we shall describe an "arrow" in E 3 by giving 
its starting point p and the change, or vector v, necessary to reach its end 
point p + v. Strictly speaking, v is just a point of E 3 . 

2.1 Definition | A tangent vector \ p to E consists of two points of E 3 : 
its vector part v and its point of application p. 

We shall always picture \ p as the arrow from the point p to the point 
p + v. For example, if p = (1,1,3) and v = (2,3,2), then \ p runs 
from (1,1,3) to (3,4,5) as in Fig. 1.1. 

We emphasize that tangent vectors are equal, \ p = w g , if and only if 
they have the same vector part, v == w, and the same point of application, 
p = q. Tangent vectors v,, and v g with the same vector part, but differ- 
ent points of application, are said to be parallel (Fig. 1.2). It is essential 
to recognize that \ p and \ q are different tangent vectors if p ^ q. In physics 
the concept of moment of a force shows this clearly enough: The same 
force v applied at different points p and q of a rigid body can produce 
quite different rotational effects. 

f A consequence is the identity / = f(x, y, z). 

% The term "tangent" in this definition will acquire a more direct geometric mean- 
ing in Chapter IV. 



Sec. 2] 



TANGENT VECTORS 



E 3 




p + v = (3, 4, 5) 



P= (1,1,3) 



FIG. 1.1 

2.2 Definition Let p be a point of E 3 . The set ^(E 3 ) consisting of all 
tangent vectors that have p as point of application is called the tangent 
space of E 3 at p (Fig. 1.3). 

We emphasize that E 3 has a different tangent space at each and every 
one of its points. 

Since all the tangent vectors in a given tangent space have the same 
point of application, we can borrow the vector addition and scalar multi- 
plication of E 3 to turn T P (E Z ) into a vector space. Explicitly, we define 
Vp _j_ Wp to be (v + w) p , and if c is a number we define c(v p ) to be (cv)„. 
This is just the usual "parallelogram law" for addition of vectors, and scalar 
multiplication by c merely stretches a tangent vector by factor c— reversing 
its direction if c < (Fig. 1.4). 

These operations on each tangent space T P (E 3 ) make it a vector space 
isomorphic to E 3 itself. Indeed it follows immediately from the definitions 
above that, for a fixed point p, the function v -» \ p is a linear isomorphism 





FIG. 1.3 



CALCULUS ON EUCLIDEAN SPACE [Chap. I 



(-**)«■ 




FIG. 1.4 



from E to ^(E ) — that is, a linear transformation which is one-to-one 
and onto. 

A standard concept in physics and engineering is that of force field. The 
gravitational force field of the earth, for example, assigns to each point 
of space a force (vector) directed at the center of the earth. 

2.3 Definition A vector field V on E 3 is a function that assigns to each 
point p of E 3 a tangent vector F(p) to E 3 at p. 

Roughly speaking, a vector field is just a big collection of arrows, one 
at each point of E 3 . 

There is a natural algebra of vector fields. To describe it, we first re- 
examine the familar notion of addition of real-valued functions / and g. 
It is possible to add/ and g because it is possible to add their values at each 
point. The same is true of vector fields V and W. At each point p, their 
values V(p) and W(p) are in the same vector space— the tangent space 
^(E 3 )— hence we can add V(p) and W(p). Consequently we can add V 
and W by adding their values at each point. The formula for this addition 
is thus the same as for addition of functions, 

(V+W)(p) = V(p) + W(p). 

This scheme occurs over and over again. We shall call it the pointwise 
principle: If a certain operation can be performed on the values of two 
functions at each point, then that operation can be extended to the func- 
tions themselves; simply apply it to their values at each point. 

For example, we invoke the pointwise principle to extend the operation 
of scalar multiplication (on the tangent spaces of E 3 ). If / is a real-valued 
function on E 3 and V is a vector field on E 3 , then fV is defined to be the 



Sec. 2] 



TANGENT VECTORS 



vector field on E 3 such that 

(fV)(p) =/( P )F(p) for all p. 

Our aim now is to determine in a concrete way just what vector fields 
look like. For this purpose we introduce three special vector fields which 
will serve as a "basis" for all vector fields. 

2.4 Definition Let U h £/ 2 , and U 3 be the vector fields on E 3 such that 

CTi(p) = (1,0,0), 

tfi(p) = (0,1,0), 

U z (p) = (0,0, l) p 

for each point p of E 3 (Fig. 1.5). We call U\, Ui, U 3 — collectively — the 
natural frame field on E . 

Thus Ui (i = 1, 2, 3) is the unit vector field in the positive x» direction. 

2.5 Lemma If V is a vector field on E 3 , there are three uniquely deter- 
mined real-valued functions vi, v iy v 3 on E 3 such that 

V = villi + V2U2 + vzUz. 

The functions Vi, v 2 , v 3 are called the Euclidean coordinate functions of V. 

Proof. By definition, the vector field V assigns to each point p a tangent 
vector V(p) at p. Thus the vector part of F(p) depends on p, so we write 
it 0>i(p), ^(p), V3(p)). (This defines V\, V2, and vz as real-valued functions 
on E 3 . ) Hence 

V(p) = (vi(p),v 2 (p),v z (p)) p 

= »i(p) (1,0,0), + t*(p) (0,1,0), + vt(p) (0,0,1), 

= vi(p)C/i( P ) + y 2 (p)f/ 2 (p) + v 3 (p)Uz(p) 

for each point p (Fig. 1.6). By our (pointwise principle) definitions this 
means that the vector fields V and X) v <Ui have the same (tangent vector) 



value at each point. Hence V = ^ ViUi 

This last sentence uses two of our stand- 
ard conventions: X) y *^» means sum over 
i = 1, 2, 3; the Halmos symbol (|) indi- 
cates the end of a proof. 

The tangent-vector identity (ai, a?., a 3 ), 
= 23°*^«(p) appearing in this proof will 
be used very often. 

Computations involving vector fields x i 
may always be expressed in terms of their 



I 




FIG. 1.5 



10 



CALCULUS ON EUCLIDEAN SPACE 



[Chap. I 



/ ' v 3 (p)Ui(-p) 7\ 




«*(P)0|(p) 



FIG. 1.6 

Euclidean coordinate functions. For example, addition, and multiplication 
by a function, are expressed in terms of coordinates by 

^ViUi + ^WiUi = £ (Vi + Wi)Ui 

Since this is differential calculus, we shall naturally require that the various 
objects we deal with be differentiate. A vector field V is differentiable 
provided its Euclidean coordinate functions are differentiable (in the sense 
of Definition 1.3). From now on, we shall understand "vector field" to 
mean "differentiable vector field." 



EXERCISES 

1. Let v = (-2, 1, -1) and w = (0, 1, 3). 

(a) At an arbitrary point p, express the tangent vector 3v p - 2w p as 
a linear combination of Ui(p), *7 2 (p), Uz(p). 

(b) For p = (1, 1, 0), make an accurate sketch showing the four 
tangent vectors v p , w p , — 2v p , and v p + w p . 

2. Let V = xU x + yU 2 and W = 2x 2 U 2 - ZJ%. Compute the vector field 
W — xV, and find its value at the point p = ( — 1, 0, 2). 

3. In each case, express the given vector field V in the standard form 

£ Villi. 

(a) 2z 2 £/ 1 = 7V + xyU*. 

(b) V(p) = (p lt p 3 - p u 0) p for all p. 

(c) V = 2(xUi + yUt) - x{U x - y 2 U 3 ). 



Sec. 3] DIRECTIONAL DERIVATIVES 11 

(d) At each point p, 7(p) is the vector from the point (pi, p 2 , pz) to 
the point (1 + p h p 2 p 3 , p 2 ). 

(e) At each point p, 7(p) is the vector from p to the origin. 

4. If V = y Ui — x Uz and W = x 2 U\ — zUz, find functions / and g such 
that the vector field fV + gW can be expressed in terms of £/ 2 and Uz 
only. 

5. Let 7x = C/i - xt/ 3 , 7 2 = £/ 2 , and 7 3 = xt/i + C/ 3 . 

(a) Prove that the vectors Fi(p), F 2 (p), F 3 (p) are linearly inde- 
pendent at each point of E 3 . 

(b) Express the vector field xU\ + yVi + zllz as a linear combination 
of 7i, 7 2 , 7,. 



3 Directional Derivatives 

Associated with each tangent vector \ p to E is the straight line t — > p + iv 
(see Example 4.2). If / is a differentiable function on E 3 , then t — >/(p + t\) 
is an ordinary differentiable function on the real line. Evidently the deriva- 
tive of this function at t = tells the initial rate of change of / as p moves 
in the v direction. 

3.1 Definition Let / be a differentiable real -valued function on E , and 
let v p be a tangent vector to E . Then the number 

y P [f] =|(/(p + «v))|« 

is called the derivative of f with respect to v p . 

This definition appears in elementary calculus with the additional 
restriction that v p be a unit vector. Even though we do not impose this 
restriction, we shall nevertheless refer to v p f/] as a directional derivative. 

For example, we compute v p f/]for the function/ = x 2 yz, with p = (1, 1, 0) 
and v = (1,0, -3). Then 

p + tv = (1, 1,0) + *(1,0, -3) = (1 + t, 1, -30 

describes the line through p in the v direction. Evaluating / along this 
line, we get 

/(p + tv) = (1 + 2 -l- (-3*) = -St - Qt 2 - St\ 
Now 

|(Ap + «v)) = -3- 12t-9t 2 ; 



--■'(./ 



12 CALCULUS ON EUCLIDEAN SPACE [Chap. I 

hence at t = 0, we find \ p [f] = — 3. Thus, in particular, the function / is 
(initially) decreasing as p moves in the v direction. 

The following lemma shows how to compute \ p \f] in general, in terms 
of the partial derivatives of / at the point p. 

3.2 Lemma If y p = (vi, v 2 , v*) p is a tangent vector to E 3 , then 

▼»[/] = £ v t |£ ( P ). 

Proof. Let p = (pi,p2,p3) ; then 

p + t\ = (pi + tv h p 2 + tv*, p 3 + tv z ). 
We use the chain rule to compute the derivative at t = of the function 

/(p + tv) = /(pi + tvi, p 2 + tv 2 , p 3 + to*). 
Since 

d 



,. (p% + to.-) = v», 



we obtain 



v p [/] = j t (/(p + *v)) |« = £ ^ (p)».-. I 

Using this lemma, we recompute v p [/] for the example above. Since 
/ = x 2 yz, we have 

Thus at the point p = (1,1,0), 

|( P ) = 0> |(p)=0, and |(p) = l. 

Then by the lemma, 

v,[/]= + 0+ (-3)1 = -3, 

as before. 

The main properties of this notion of derivative are given in Theorem 3.3. 

3.3 Theorem Let / and g be functions on E 3 , v p and w p tangent vectors, 
a and b numbers. Then 

(1) (av p + bw,)[/l = av p [/] + bw p [f\. 

(2) v p [af + bg] = av p [/] + ftvpfo]. 

(3) v,\fg] = y P Lf]-^(p) +/(p)-v,fe]. 



Sec. 3] DIRECTIONAL DERIVATIVES 13 

Proof. All three properties may be deduced easily from the preceding 
lemma. For example, we prove (3). By the lemma, if v = {v\,vi,vz), then 

v,[/<7l = & ^ (P). 



But 



Hence 



dxi dXi dXi 



-(Z*£(P>><P>+/<P>(2>.*;<r>) 

= v,l/l-«f(p) +/(p) •▼,!»]. I 

The first two properties in the preceding theorem may be summarized 
by saying that \ p [f\ is linear in v p and in /. The third property, as its proof 
makes clear, is essentially just the usual Leibniz rule for differentiation of 
a product. No matter what form differentiation may take, it will always have 
suitable linear and Leibnizian properties. 

We now use the pointwise principal to define the operation of a vector 
field V on a function f. The result is the real-valued function V[f] whose 
value at each point p is the number V(p)\f\ — that is, the derivative of/ 
with respect to the tangent vector F(p) at p. This process should be no 
surprise, since for a function / on the real line, one begins by defining the 
derivative of fata point — then the derivative function df/dx is the function 
whose value at each point is the derivative at that point. Evidently the 
definition of V\f] is strictly analogous to this familiar process. In particular, 
if Ui, U%, U% is the standard frame field on E 3 , then Ui[f] = df/dxi. This 
is an immediate consequence of Lemma 3.2. For example, ZJ\ (p) = (l,0,0) p ; 
hence 

Ui(p)\f\ = | (/(pi + t, p*, p 3 ))|«=o, 

which is precisely the definition of (df/dxi) (p). This is true for all points 
P = (Pi,P2,Ps); hence Uitf] = df/dxi. 

We shall use this notion of directional derivative more in the case of 
vector fields than for individual tangent vectors. 

3.4 Corollary If V and W are vector fields on E , and /, g, h are real- 
valued functions, then 



14 CALCULUS ON EUCLIDEAN SPACE [Chop. I 

(1) (fV + gW)[h] = fV[h] + gW[h]. 

(2) V[af + bg] = aV[f] + bV\g], for all real numbers a and b. 

(3) V\fg]= V[f\-g + fV\g]. 

Proof. The pointwise principle guarantees that to derive these properties 
from Theorem 3.3 we need only be careful about the placement of paren- 
theses. For example, we prove the third formula. By definition, the value 
of the function V[fg] at p is V(p)\fg]. But by Theorem 3.3 this is 

V(p)\f\-g(p) +/(p)7(p)fo] = F[/](p).<7(p) +/(p)7M(p) 

= (V\f\'9+f-V\g])(p). I 

If the use of parentheses here seems extravagant, we remind the reader 
that a meticulous proof of Leibniz' formula 

— (f ) = ^l» -\- f.^ 
dx dx dx 

must consist of exactly the same shifting of parentheses. 

Note that the linearity of V[f] in V and / is for functions as "scalars" 
in the first formula in Corollary 3.4 but only for numbers as "scalars" in 
the second. This stems from the fact that/F signifies merely multiplication, 
but V\f] is differentiation. 

The identity Ui[f] = df/dxi makes it a simple matter to carry out explicit 
computations. For example, if V = xUi — y 2 U s and/ = x 2 y + z s , then 

V[f] = xUAx'y] + xUA* 3 } - y 2 U s [x 2 y] - y 2 U 3 [z 3 } 

= x(2xy) + - - y 2 (Sz 2 ) = 2x 2 y - 3y 2 z 2 . 

3.5 Remark Since the subscript notation \ p for a tangent vector is 
somewhat cumbersome, from now on we shall frequently omit the point 
of application p from the notation. This can cause no confusion, since v 
and w will always denote tangent vectors, and p and q points of E 3 . In 
many situations (for example, Definition 3.1) the point of application is 
crucial, and will be indicated by using either the old notation v,, or the 
phrase "a tangent vector v to E 3 at p." 



EXERCISES 

1. Let \ p be the tangent vector to E 3 for which v = (2, —1, 3) and 
p = (2, 0, —1). Working directly from the definition, compute the 
directional derivative v p [/], where 
(a) / = yz. (b) / = x\ (c) / = e cos y. 



Sec. 4] CURVES IN E 3 15 

2. Compute the derivatives in Ex. 1 using Lemma 3.2. 

3. Let V = y 2 Ux - xU z , and let/ = xy,g = 2 3 . Compute the functions 

(a) 7[fl. (c) 7[M (e) F[f + </ 2 ]. 

(b) V\g\. (d) /7fe] - flf7[/]. (f) V[V\J]l 

4. Prove the identity 7=2 Vfo]l7i, where xi, a*, x 3 are the natural 
coordinate functions. (Hint: Evaluate V = 2 ViU t on xy.) 

5. If 7[fl = WLfl for every function / on E 3 , prove that V = W. 



4 Curves in E 3 

Let I be an open interval in the real line R. We shall interpret this liberally 
to include not only the usual type of open interval a < t < b (a, b real 
numbers), but also the types a < t (a half -line to +«),*< 6 (a half -line 
to — oo ), and also the whole real line. 

One can picture a curve in E 3 as a trip taken by a moving point a. At 
each "time" t in some open interval, a is located at the point 

a(t) = (ai(0,« 2 (0, «s(0) 

in E 3 . In rigorous terms then, a is a function from / to E 3 , and the real- 
valued functions a u a 2 , a 3 are its Euclidean coordinate junctions. Thus we 
write a = (ai,a2,a8), meaning, of course, that 

a(jt) = (ai(0,« 2 (0,«3(0) 

for all •* in the interval I. We define the function a to be differentiate 
provided its (real-valued) coordinate functions are differentiate in the 
usual sense. 

4.1 Definition A curve in E 3 is a differentiate function a: I —*■ E from 
an open interval I into E . 

We shall give several examples of curves, which will be used in Chapter 
II to experiment with results on the geometry of curves. 

4.2 Example 

(1) Straight line. A line is the simplest type of curve in Euclidean space; 
its coordinate functions are linear (in the sense t — > at + 6, not in the 
homogeneous sense t -^ at). Explicitly, the curve a: R -* E 3 , such that 

a (t) = p + tq = (pi + tqi, p 2 + tq 2 , p 3 + tqt) (q ^ 0) 

is the straight line through the point p = a(0) in the q direction. 

(2) Helix. (Fig. 1.7). The curve t -> (a cos t, a sin t, 0) travels around 



w 



CALCULUS ON EUCLIDEAN SPACE 



[Chap. I 





FIG. 1.7 



FIG. 1.8 



a circle of radius a > in the xy plane of E 3 . If we allow this curve to rise 



(or fall) at a constant rate, we obtain a helix a: R 
formula 



E , given by the 



a {t) = (a cos t, a sin t, bt) 

where a > 0, 6 ^ 0. (We shall always use the term helix to mean right 
circular helix. ) 

(3) Let 

<x(t) = (2 cos 2 /, sin 2t, 2 sin t) for < t < w/2. 

This curve a has a noteworthy property: Let C be the cylinder in E 3 con- 
structed on the circle in the xy plane with center at (1,0,0) and radius 1. 
Then a follows the route sliced from C by the sphere S with radius 2 and 
center at the origin (Fig. 1.8). 

(4) The curve a: R — > E 3 such that 

a(t) = (e\e~\y/2t) 

shares with the helix in (2) the property of rising constantly. However, it 
lies over the hyperbola xy = 1 in the xy plane instead of a circle. 

(5) The curve a: R -► E 3 such that 

a(t) = (St - t\St 2 ,3t + t s ). 

If the coordinate functions of a curve are simple enough, the shape the 
curve has in E 3 can be found, at least approximately, by the brute-force 
procedure of plotting points. We could get a reasonable picture of this curve 
for ^ t ^ 1 by computing a(t) for t = 0, Ho, H, Ho, 1. 

If we visualize a curve a in E 3 as a moving point, then at every time t 



Sec. 4] CURVES IN E 3 17 

there is a tangent vector at the point a(t) which gives the instantaneous 
velocity of a at that time. 

4.3 Definition Let a: / — ► E be a curve in E with a = (a 1} an, 013). For 
each number t in /, the velocity vector of a at t is the tangent vector 



«'(0 -(£<o,^<0.^(«) 



dt dt dt }a{t) 

at the point a(t) in E a (Fig. 1.9). 

We interpret this definition geometrically as follows. The derivative 
at t of a real-valued function / on R is given by 

dt At-o At 

This formula still makes sense if/ is replaced by a curve a = (an, 0:2,0:3). 
In fact, 

<«(« + M) - «(()) = • " (t + M) ~ aM 



c- 



At" V At 

a 2 « + At) - a 2 (t) a t (t + At) - a»(t) \ 
At ' At )' 

This is the vector from a(t) to a(t + At), scalar multiplied by l/At (Fig* 
1.10). 

Now as At gets smaller, a(t -"- Af) approaches a(0> and in the limit as 
A£ — > 0, we get a vector tangent to the curve a at the point a(t), namely, 
(dai/dt(t), da 2 /dt(t), da 3 /dt(t)). As the figure suggests, the point of 
application of this vector must be the point a(t). Thus the standard limit 
operation for derivatives gives rise to our definition of the velocity of a 
curve. 



<*'(*) 




FIG. 1.9 



18 



CALCULUS ON EUCLIDEAN SPACE 
1 



[Chap. I 




(a(t + At) - a(t)) 



FIG. 1.10 



An application of the identity 

to the velocity vector a (t) at t yields the alternative formula 

«'(*) = E^(0^(«(0). 
For example, the velocity of the straight line a(t) = p + tq is 

«'(0 = (qi,qt,qi)aw= q«<«). 

The fact that a is straight is reflected in the fact that all its velocity vectors 
are parallel; only the point of application changes as t changes. 



For the helix 



the velocity is 



a(t) = (a cos t, a sin t, bt), 



a (0 = (—a sin t, a cos t, b) 



a(t)- 



The fact that the helix rises constantly is shown by the constancy of the 
z coordinate of a (t). 

Given a curve a, one can construct many new curves which follow the 
same route as a, but travel at different speeds. 

4.4 Definition Let / and J be open intervals in the real line R. Let a: 
I -+ E be a curve and let h: J -► / be a differentiate (real-valued) func- 
tion. Then the composite function 

= a(h): J-+E 3 
is a curve called the reparametrization of a by h. 

At each time s in the interval J, the curve p is at the point (*) = « (A (a) ) 
reached by the curve a at time A(s) in the interval / (Fig. 1.11). Thus 
does follow the route of a, but fi generally reaches a given point on the route 
at a different time than a does. In practice, to compute the coordinates of 
0, one simply substitutes t = h(s) in the coordinates ai(t), a 2 (t), a »(t) of 



Sec. 4] CURVES IN E 3 19 




FIG. 1.11 

a. For example, suppose a(t) = (\/t, ty/l, 1 — t) on J: < t < 4. If 
h(s) = s 2 on J: < s < 2, then 

0(s) = a(Ms)) = «(s 2 ) = (s, s\ 1 - s 2 ). 

Thus the curve a: I -> E 3 has been reparametrized by to to yield the curve 
0: J->E 3 . 

The following lemma relates the velocities of a curve and of a repara- 
metrization. 

4.5 Lemma If /3 is the reparametrization of a by h, then 

/?'(«) = (dh/ds)(s)a(h(s)). 
Proof. If a = (ai, a2, 03), then 

j8(s) = a(ft(*)) = (ai(A(s)),a,(A(*)),a,(A(«)). 

Using the "prime" notation for derivatives, the chain rule for a composi- 
tion of real-valued functions / and g reads (g(J)Y = g'(J)'f- Thus in tne 
case at hand, we obtain 

aM'is) = ai '(h(s))-h'(s). 

By the definition of velocity, this yields 

ft {*) = «(*)'(«) 

= (a/ (*(«)) •*'(«),«* (*(«))•* («),«» (*(•))•* («)) 
= *'(*)«'(*(«))■ I 

According to this lemma, to obtain the velocity of a reparametrization 
of a by h, first reparametrize a by h, then scalar multiply by the derivative 

of ft. 

Since velocities are tangent vectors, we can take the derivative of a 
function with respect to a velocity. 

4.6 Lemma Let a be a curve in E 3 and let / be a differentiable function 
on E 3 . Then 

«'(0W - d ^ }1 (0. 



20 



CALCULUS ON EUCLIDEAN SPACE 



[Chap. I 



Proof. Since 






we conclude from Lemma 3.2 that 



«'(Wi = Z^(«(0)^«). 



dXi 



dt 



But the composite function /(a) may be written /(ai,a 2 ,a 3 ), and the 
chain rule then gives exactly the same result for the derivative of f(a). | 

By definition, a' (Of/] is the rate of change of / along the line through 
a(t) in the a (t) direction (Fig. 1.12). (If a (t) * 0, this is the tangent 
line to a at a (t) ; see Exercise 9.) The lemma shows that this rate of change 
is the same as that of / along the curve a itself. 

Since a curve a: I — ► E 3 is a function, it makes sense to say that a is one- 
to-one; that is, a(t) = a(ti) only if t — h. Another special property of 
curves is periodicity: A curve a: R -> E 3 is periodic if there is a number 
p > such that a(t + p) = a(t) for all t— and the smallest such number 
p is then called the period of a. 

From the viewpoint of calculus, the most important condition on a 
curve a is that it be regular, that is, have all velocity vectors different from 
zero. Such a curve can have no corners or cusps. 

The following remarks about curves (offered without proof) are not an 
essential part of our exposition, but will be of use in Chapter IV. We con- 
sider, in the case of the plane E 2 , another familiar way to formulate the 
concept of "curve." If / is a differentiable real-valued function on E 2 , let 

C:f=a 

be the set of all points p in E 2 such that /(p) = a. Now if the partial 
derivatives df/dx and df/dy are never simultaneously zero at any point of 
C, then C consists of one or more separate "components" which we shall 




^ 


/. 


\\ 


// 


\s 


// 


\\ 


/ / 


\ s 


/ 1 


\ \ 


/ I 


^21 \ 


/ fc. 


j • 


\ 1 


/ / 


N \ 


1 / 


\ \ 


1 / 


\ \ 


/ / 


\ \ 


// 


\\ 


// 


>X 


V 


V 



FIG. 1.12 



FIG. 1.13 



Sec. 4] CURVES IN E 3 21 

call Curves^ Thus, for example, C: x + y 2 = r 2 is the circle of radius r 
centered at the origin of E 2 , and the hyperbola C: x — y 2 — r 2 splits into 
two Curves ("branches") d and C 2 as in Fig. 1.13. 

Every Curve C is the route of many regular curves a, called parametriza- 
tions of C. If C is a closed Curve, then it has a periodic parametrization 
a: R — > C. For example, the curve 

a(t) = (r cos t, r sin t) 

is a well-known periodic parametrization of the circle given above. If C 
is a Curve that is not closed (sometimes called an arc), then every para- 
metrization £: / — > C is one-to-one. For example, 

/8(0 = (r cosh t, r sinh £) 

parametrizes the branch, x > 0, of the hyperbola given above. 



EXERCISES 

1. Compute the velocity vector of curve (3) in Example 4.2 for arbitrary 

t, and at t = x/4. 

2. Plot curve (5) in Example 4.2 by the method there suggested. On 
the sketch show the velocity vectors at t = 0, |, 1. 

3. Find the coordinate functions of the curve = a(h), where a is curve 
(3) in Example 4.2 and h is the function on J: < s < 1 such that 
h(s) = sin -1 s. 

4. Find the (unique) curve a such that a(0) = (1,0, —5) and a (t) = 
(t\t,e<). 

5. Find a straight line passing through the points (1, —3, —1) and 
(6, 2, 1). Does this line meet the line through the points (—1, 1, 0) 
and (-5, -1, -1)? 

6. Deduce from Lemma 4.6 that in the definition of directional derivative 
(Definition 3.1) the straight line t — ► p + tv may be replaced by any 
curve a with initial velocity v p , that is, such that a(0) = p and a (0) = 
v p . 

7. (Continuation). Show that the curves given by (/, 1 + t 2 , t), (sin t, cos t, 
t), and (sinh t, cosh t, t) all have the same initial velocity \ p . 
If / = x 2 — y + z, compute v p \f] by evaluating / on each of the curves. 

8. Let h(s) = log s on J: s > 0. Reparametrize curve (4) in Example 4.2 

t In this section (only) we use a capital C to distinguish this notion from a curve 
a:/^E 3 . 



22 CALCULUS ON EUCLIDEAN SPACE [Chap. I 

using h. Check the equation in Lemma 4.5 in this case by computing 
each side separately. 

9. For a fixed t, the tangent line to a regular curve a at a (t) is the straight 
line u — ► a(t) + ua'(t), where we delete the point of application of 
a (t). Find the tangent line to the helix a(t) = (2 cos t, 2 sin t, t) at 
the points <*(0) and a(x/4). 

10. Sketch the following Curves in E 2 and find parametrizations for each. 

(a) (7:4/ + y 2 = 1. (c) C: y = e x . 

(b) C: Zx + 4y = 1. (d) C: z 2/3 + t/ 2/3 = 1, x > 0, j, > 0. 



5 1 -Forms 

If / is a real-valued function on E 3 , then in elementary calculus one defines 
the differential of / to be 

df=~dx+ d ldy+ d /dz. 
dx dy * dz 

It is not always made clear exactly what this formal expression means. 
In this section we give a rigorous treatment using the notion of 1-form, 
and these will tend to appear at crucial moments in our later work. 

5.1 Definition A 1-form <j> on E 3 is a real-valued function on the set of 
all tangent vectors to E 3 such that <j> is linear at each point, that is, 

4>(a\ + 6w) = cuf>(\) + b<f>(w) 

for any numbers a, b and tangent vectors v, w at the same point of E 3 . 

We emphasize that for every tangent vector v to E 3 , a 1-form <f> defines 
a real number <f> (v) ; and for each point p in E 3 , the resulting function 
<j> p : T P (E ) — »• R is linear. [Thus at each point p, <f> p is an element of the 
d-wl space of ^(E 3 ). In this sense the notion of 1-form is dual to that of 
vector field.] 

The sum of 1 -forms <t> and \p is defined in the usual pointwise fashion 

(<t> + ^) (v) = <£(v) + ^(v) for all tangent vectors v. 

Similarly if / is a real -valued function on E 3 and </> is a 1-form, then fy> 
is the 1-form such that 

(fo)(vp) =/(p)*(v,) 
for all tangent vectors v p . 

There is also a natural way to evaluate a 1-form <j> on a vector field V to 
obtain a real-valued function 4>(V): At each point p the value of <f>(V) 



Sec. 51 1 -FORMS 23 

is the number <f>(V(p)). Thus a 1-form may also be viewed as a machine 
which converts vector fields into real-valued functions. If <f> (V) is differenti- 
able whenever V is, we say that <£ is differentiable. As with vector fields, 
we shall always assume that the 1 -forms we deal with are differentiable. 

A routine check of definitions shows that <j>(V) is linear in both and V; 
that is, 

*(fV + gW) = f<t>(V) +g<t>{W) 
and 

(ft + 0*)(F) = f<t>(V) +g + (V) 

where / and g are functions. 

Using the notion of directional derivative, we now define a most impor- 
tant way to convert functions into 1 -forms. 

5.2 Definition If / is a differentiable real-valued function on E 3 , the 
differential df of / is the 1-form such that 

df(v p ) = y p [f] for all tangent vectors y p . 

In fact, df is a, 1-form, since by definition it is a real-valued function on 
tangent vectors, and by (1) of Theorem 3.3 it is linear at each point p. 
Clearly df knows all rates of change of / in all directions on E 3 , so it is not 
surprising that differentials are fundamental to the calculus on E 3 . 

Our task now is to show that these rather abstract definitions lead to 
familiar results when expressed in terms of coordinates. 

5.3 Example 1 -Forms on E 3 . (1) The differentials dx lf dx 2 , dx 3 of the 
natural coordinate functions. Using Lemma 3.2 we find 

dxi(\ p ) = v p [ Xi ] = Yj>i jt 1 (p) = Yji s n = v i 

j OXj j 

where «, 7 is the Kronecker delta (0 if i 9± j, 1 if i = j). Thus the value of dx { 
on an arbitrary tangent vector v p is the ith coordinate Vi of its vector -part — and 
does not depend at all on the point of application p. 

(2) The 1-form f = fi dxi + / 2 dx 2 + f 3 dx 3 . Since dxi is a 1-form, our 
definitions show that ^ is also a 1-form for any functions f u / 2 , / 3 . The value 
of ^ on an arbitrary tangent vector v p is 

*(▼») = (£/«**)(▼,) = Hfi(p)dx i (y) = E/<(p)»* 

The first of these examples shows that the 1 -forms dx u dx 2 , dx 3 are the 
analogues for tangent vectors of the natural coordinate functions xi, z 2 , x 3 
for points. Alternatively, we can view dx u dx 2 , dx 3 as the "duals" of the 
natural unit vector fields U u U 2 , U 3 . In fact, it follows immediately from 
(1) above that the function dXi(Uj) has the constant value 5 t7 . 



24 CALCULUS ON EUCLIDEAN SPACE [Chap. I 

We shall now show that every 1-form can be written in the concrete 
manner given in (2 ) above. 

5.4 Lemma If <£ is a 1-form on E , then <f> = 52 fi dxi, where / t = <f>(Ui). 
These functions /i, / 2 , / 3 are called the Euclidean coordinate functions of <f>. 

Proof. By definition a 1-form is a function on tangent vectors; thus <f> 
and 52 /*' dxi are equal if and only if they have the same value on every 
tangent vector v p = 52 ViUi(p). In (2) of Example 5.3 we saw that 

(2/.-dx.-)(vp) = 52/*(pH- 
On the other hand, 

*(v,) =0(X>;£A(p)) = 5>**»(^(p)) = I><A(p) 

since /» = <j>(Ui). Thus <f> and 52 fidxt do have the same value on every 
tangent vector. | 

This lemma shows that a 1-form on E is nothing more than an expres- 
sion fdx + g dy + h dz, and such expressions are now rigorously defined 
as functions on tangent vectors. Let us now show that the definition of 
differential of a function (Definition 5.2) agrees with the informal defini- 
tion given at the start of this section. 

5.5 Corollary If / is a differentiate function on E 3 , then 

Proof. The value of 52 (df/dxi) dxi on an arbitrary tangent vector 
v p is 52 (df/dXi)(p)vi. By Lemma 3.2 df{\ p ) = \ p [f] is the same. Thus 
the 1 -forms df and 52 W/dx,-) dxi are equal. | 

Using either this result or the definition of d, it is immediate that 

dV+g) = df + dg. 

Finally we determine the effect of d on products of functions and on compo- 
sitions of functions. 

5.6 Lemma Let fg be the product of differentiate functions / and g 
on E 3 . Then 

d(fg) = gdf + fdg. 
Proof. Using Corollary 5.5, we obtain 

«*>-E'4?*<-z(£;. + /i->. 



Sec. 5] 1-FORMS 25 

5.7 Lemma Let /: E 3 — > R and fc: R — > R be differentiable functions, 
so the composite function h (/) : E 3 — ► R is also differentiable. Then 

<*(*(/)) =h'(f)df. 

Proof. (The prime here is just the ordinary derivative, so h (/) is again 
a composite function, from E 3 to R.) The usual chain rule for a composite 
function such as h (/) reads 

m» . kvff 

dXi dXi 

Hence . 

d(Hf)) = £ *<^ <&* = £ fc'(/> ^ <fe, = h'(f)df. | 

To compute df for a given function / it is almost always simpler to use 
these properties of d rather than substitute in the formula of Corollary 
5.5. Then from df we immediately get the partial derivatives of / and, in 
fact, all its directional derivatives. For example, suppose 

/= (x 2 - l)y + Q/ 2 + 2)z. 

Then by Lemmas 5.6 and 5.7, 

df = (2x dx)y + (x 2 - 1) dy + (2y dy)z + (y 2 + 2) dz 

= 2xy dx + (x 2 + 2 yz - \) dy + (y 2 + 2) dz 

d//dx d//3y d//da 

Now use the rules above to evaluate this expression on a tangent vector 
v . The result is 

v,[/l = df(y p ) = 2pip 2 fi + (pi 2 + 2p 2 p 3 - l)w« + (P2 2 + l)w». 



EXERCISES 

1. Let v = (1, 2, —3) and p = (0, —2, 1). Evaluate the following 
1 -forms on the tangent vector v p . 

(a) y 2 dx. (b) zdy — y dz. (c) (z 2 — \)dx — dy + z 2 dz. 

2. If <t> = 52 /f dx» an d V = 22 v *^«> show that the 1-form </> evaluated 
on the vector field V is the function 4>(V) = 22 /»' y <- 

3. Evaluate the 1-form <£ = x 2 dx — y 2 dz on the vector fields 
V = xU x + ?/t/ 2 + z£/ 3 , 

TF = xy{U, - U t ) + y«(tfi - Ut), and (l/z)F + (l/y)W. 



26 CALCULUS ON EUCLIDEAN SPACE [Chap. I 

4. Express the following differentials in terms of df: 

(a) d(f). (b) d(VJ), where/ > 0. (c) d(log(l +/ 2 )). 

5. Express the differentials of the following functions in the standard 
form ^fidxi. 

(a) {x 2 + y 2 + i) m . (b) tan^O//*). 

6. In each case compute the differential of / and find the directional de- 
rivative v p [/], for v p as in Ex. 1. 

(a) / = xy 2 — yz. (b) / = xe yz . (c) / = sin {xy) cos {xz). 

7. Which of the following are 1 -forms? In each case <f> is the function on 
tangent vectors such that the value of <f> on (vi, v 2 , v 3 ) p is 

(a) Vi — vs. (c) vtps + v 2 pi. (e) 0. 

{b) Pl -p z . {d) v p [x 2 + /}. (f) (pi) 2 . 
In case <j> is a 1-form, express it as 23 /*" ^»- 

8. Prove Lemma 5.6 directly from the definition of d — without using 
Corollary 5.5. 

9. A 1-form is zero at a point p provided 4>{v P ) = for all tangent 
vectors at p. A point at which its differential df is zero is called a 
critical point of the function /. Prove that p is a critical point of / 
if and only if 

d l ( p ) = d l ( p ) = d l ( p ) = o. 
dx p/ by p bz vp/ 

Find all critical points of / = (1 — x 2 )y + (1 — y )z. 

{Hint: Find the partial derivatives of / by computing df. ) 

10. {Continuation) . Prove that the local maxima and local minima of / 
are critical points of /. (/ has a local maximum at p if /(q) ^ /(p) 
for all q near p.) 

11. It is sometimes asserted that df is the linear approximation of A/. 

(a) Explain the sense in which {df) (v p ) is the linear approximation 
of/(p + v) -/(p). 

(b) Compute the exact and approximate values of / (0.9, 1.6, 1.2) — 
/(l, 1.5, 1), where/ = x 2 y/z. 



6 Differential Forms 

The 1 -forms on E 3 are part of a larger system called the differential forms 
on E 3 . We shall not give as rigorous an account of differential forms as we 
did of 1 -forms, for o"r use of the full system will be limited to Section 8 



Sec. 6] DIFFERENTIAL FORMS 27 

of Chapter II. Roughly speaking, a differential form on E 3 is an expression 
obtained by adding and multiplying real-valued functions and the differ- 
entials dxi, dx2, dxz of the natural coordinate functions of E . These two 
operations obey the usual associative and distributive laws; however, the 
multiplication is not commutative. Instead it obeys the 

alternation rule: dxidxj = —dxjdxi (1 ^ i, j ^3). 

This rule appears — although rather inconspicuously — in elementary 
calculus (see Exercise 9). 

A consequence of the alternation rule is the fact that "repeats are zero," 
that is, dxi dxi = 0, since if i — j the alternation rule reads 

If each summand of a differential form contains pdxi's (p = 0, 1, 2, 3), 
the form is called a p-form, and is said to have degree p. Thus, shifting to 
dx,dy,dz, we find 

A 0-form is just a differentiable function /. 

A 1-form is an expression fdx -f g dy + hdz, just as in the preceding 

section. 
A 2 -form is an expression / dx dy + g dx dz + h dy dz. 
A 3 -form is an expression / dx dy dz. 

We already know how to add 1 -forms: simply add corresponding coef- 
ficient functions. Thus, in index notation, 

2 fi dxi -f X 9i dxi = 53 (fi + ffi) dxi. 

The corresponding rule holds for 2-forms or 3-forms. 

On three-dimensional Euclidean space, all p-forms with p > 3 are zero. 
This is a consequence of the alternation rule, for a product of more than 
three dxi's must contain some dx t twice, but repeats are zero, as noted 
above. For example, dxdydxdz = —dxdxdydz = 0, since dxdx = 0. 
As a reminder that the alternation rule is to be used, we denote this multi- 
plication of forms by a wedge a . (However, we do not bother with the 
wedge when only products of dx, dy, dz are involved.) 

6.1 Example Computation of wedge products. (1) Let 

tf> = xdx — y dy and ty = z dx + x dz. 
Then 
4> a ^ = (x dx — y dy) a (z dx + x dz) 

= xz dx dx + x 2 dx dz — yz dy dx — yx dy dz. 



28 CALCULUS ON EUCLIDEAN SPACE [Chap. I 

But 

dx dx = 0, and dy dx = — dx dy. 
Thus 

4> a \f/ = yz dx dy + x 2 dx dz — xy dy dz. 

In general, the product of two 1 -forms is a 2-form. 

(2) Let <f> and \f/ be the 1 -forms given above and let 

9 = z dy. 
Then 

a $ a 4/ = yz" dy dx dy + x 2 z dy dx dz — xyz dy dy dz. 
Since dy dx dy and dy dy dz each contain repeats, both are zero. Thus 
a a \f/ = —x 2 z dx dy dz. 

(3 ) Let <f> be as above, and let 77 be the 2-form y dx dz + x dy dz. Omitting 
forms containing repeats, we find 

$ a 77 = x 2 dxdy dz — y 2 dy dx dz = (x 2 + y 2 ) dx dy dz. 

It should be clear from these examples that the wedge product of a 
p-form and a q-iovm. is a (p + q)-iorm. Thus such a product is automatically 
zero whenever p + q > 3. 

6.2 Lemma If $ and ^ are 1 -forms, then 

<f> A ^ = — ^ A 0. 

Proof. Write 

<t> = X /i <&••> ^ = £ 0i fe 

Then by the alternation rule, 

<t> A 4> — ]£/#/ cfoi dx, = - J] flfj/i cte; dxi = —4* a <f>. | 

In the language of differential forms, the operator d of Definition 5.2 
converts a 0-form / into a 1-form df. It is easy to generalize to an operator 
(also denoted by d) which converts a p form 77 into a (p + l)-form ^77: 
One simply applies d (of Definition 5.2) to the coefficient functions of 77. 
For example, here is the case p = 1 . 

6.3 Definition If <f> = ]£ f { dxi is a 1-form on E 3 , the exterior derivative 
of <f> is the 2-form d4> = ^ dfc a dx { . 

If we expand the preceding definition using Corollary 5.5, we obtain the 



Sec. 6] DIFFERENTIAL FORMS 29 

following interesting formula for the exterior derivative of 
<f> = /i dxi + / 2 dxi + fz dx z : 



dxi dx%l 



\dxi 8x2/ \dxi dx s / \ 

There is no need to memorize this formula; it is more reliable to simply 
apply the definition in each case. For example, suppose 



4> = xy dx + x dz. 



Then 



d4> = d(xy) a dx + d(x ) dz 

— (y dx + x dy) a dx + 2x dx dz 
= —x dx dy + 2x dx dz. 

It is easy to check that the general exterior derivative enjoys the same 
linearity property as the particular case in Definition 5.2; that is, 

d{a4 + 6^) = a d<t> + b d$, 

where </> and \f/ are arbitrary forms and a and 6 are numbers. 

The exterior derivative and the wedge product work together nicely. 

6.4 Theorem Let / and g be functions, <j> and \f/ 1 -forms. Then 

(1) d(fg) =dfg+fdg. 

(2) d(f<f>) = df a 4>+fd<t>. 

(3) d(4> a yp) = d<f> a yp — $ a d\p. f 

Proof. The first formula is just Lemma 5.6. We include it to show the 
family resemblance of all three formulas. The proof of the second formula 
is a simpler variant of that of the third, so we prove only the latter. 

Cose 7 . <f> = f dx, *p = g dx. Since 

4> a \p = fg dx dx = 0, 
we must show that the right side of the equation is also zero. Now 

dtf> = df a dx = -^- dy dx + -^ dz dx: 
dy dz 

hence each term of cUp a ^ has a repeated dx. Thus d$ a \f/ = 0, and simi- 
larly <£ a dip = 0. 

Case 2. <f> = f dx, \p = g dy. Using the formula for d^ computed above, 
we get 

t As usual, multiplication takes precedence over addition or subtraction, so this 
expression should be read as (dtj> A 4) — (<f> A <bf). 



30 CALCULUS ON EUCLIDEAN SPACE [Chap. I 

dtf> a yf, = iJ- dy dx + j- dz dx) a g dy 



Similarly, 



Thus 



But 



so we get 



= + -^- g dz dx dy = g -^- dx dy dz. 
dz dz 



<f> a dip = fdx a ( -? dx dy + -? dz dy ) 
\dx dz / 

= f -f dx dz dy = —f^dxdy dz. 
dz dz 



* t - <t> a df = \g£ + ^ /J dx dy dz. 
4 a $ = fg dx dy, 



d(<f> a *) = <*(#) da: dy = ^1 dz dx dy 

dz 

Hence the formula is proved in this case. 

Case 3. The general case. From cases 1 and 2 we know that the formula 
is true whenever <f> and ^ are "simple," that is, of the form / du, where u 
is x, y, or z. Since every 1-form is a sum of simple 1 -forms, the general case 
follows from the linearity of d and the distributive law for the wedge 
product. | 

One way to remember the minus sign which occurs in formula (3) of 
Theorem 6.4 is to pretend that d is a 1-form. To reach ^, d must change 
places with <f>; hence the minus sign is consistent with Lemma 6.2. 

Differential forms, and the associated concepts of wedge product and 
exterior derivative, provide a means of expressing rather complicated rela- 
tionships in a simple, methodical way. For example, as its proof shows, 
the tidy formula 

d(<f> a \f/) = d<f> a ^ — <f> a d\{/ 

involves some rather tricky relations among partial derivatives. Before 



Sec. 61 DIFFERENTIAL FORMS 31 

forms were invented, it was necessary to struggle through these relations 
in many a separate problem, but now we simply apply the general formula. 
A variety of interesting applications is given in Flanders [1]. Later on 
we shall use differential forms to express the fundamental equations of 
geometry. 



EXERCISES 

1. Let <f> = yz dx + dz, \f/ = sin z dx + cos z dy, £ = dy -+- z dz. Find the 
standard expressions (in terms of dx dy, • • • ,) for 

(a) a f, $ a £, % a <t>. (b) dxf>, dxf,, d$. 

2. Let <f> = dx/y and ^ = z dy. Check the Leibnizian formula (3) of The- 
orem 6.4 in this case by computing each term separately. 

3. For any function / show that d (df) = 0. Deduce that d{f dg) = df a dg. 

A. Simplify the following forms : 

(z)d(fdg + gdf). (c) d(fdg a g df). 

(b) d{ (/ - g)(df + dg)}. (d) d(gfdf) + d(fdg). 

5. For any three 1 -forms </>; = ^jfijdxj (1 ^ i ^ 3), prove 



01 A <f>2 A <f>3 = 



/ll /l2 /l3 

/21 /22 /23 etei cte 2 dx 3 . 

/31 /32 J33 

6. If r, #, and z are the cylindrical coordinate functions on E 3 , then 
x = r cos &, y = r sin #, z = z. Compute the volume element dx dy dz 
of E in cylindrical coordinates. (That is, express dx dy dz in terms of 
the functions r, tp, z and their differentials. ) 

7. For a 2-form 

77 = f dx dy -\- g dx dz -\- h dy dz, 

the exterior derivative dr\ is defined to be the 3-form obtained by replac- 
ing /, g, and h by their differentials. Prove that for any 1-form 
<f>,d(d<t>) = 0. 

Exercises 3 and 7 show that d 2 = 0, that is, for any form £, d(di-) = 0. 
(If £ is a 2-form, then d(d£) = 0, since its degree exceeds 3.) 

8. Classical vector analysis avoids the use of differential forms on E by 
converting 1 -forms and 2-forms into vector fields by means of the follow- 
ing one-to-one correspondences: 

£ fi dxi -^- X f&i -^- U dxi dxi — / 2 dxi dx 3 + f x dx 2 dx 3 . 



32 



CALCULUS ON EUCLIDEAN SPACE 



[Chap. I 



Vector analysis uses three basic operations based on partial differentia- 
tion: 

Gradient of a function /: 

grad/= £^-t/;. 
dXi 

Curl of a vector field V = J^ftUi: 

curl V = (°* - d A ) Ul + (* - *) U, + (* - * W 

\dx 2 dxj ^ \dx 3 dxj T \d:n dz 2 / 3 

Divergence of a vector field V — ^ /»#*: 

div 7 = V 3£. 

Prove that all three operations may be expressed by exterior deriva- 
tives as follows: 
(a)d/JiLgrad/. 

(b) If <f> J1L V, then d0 J^L curl F. 

(c) If t, J^L V, then dr, = (div 7) dx dy dz. 

9. Let / and g be real-valued functions on E 2 . Prove that 



df a dg = 



5/ 


5/ 


d£ 


ay 


d<7 


d<7 


3a; 


dy 



dx cfy. 



This formula appears in elementary calculus; show that it implies the 
alternation rule. 



7 Mappings 

In this section we discuss functions from E" to E m . If n = 3 and m = 1, 
then such a function is just a real-valued function on E 3 . If n = 1 and 
m = 3, then such a function is a curve in E 3 . Although our results will 
necessarily be stated for arbitrary m and n, we are primarily interested 
only in the three cases: 



E 2 ^E 2 



E 2 



E* 



E\ 



The fundamental observation about a function F: E" — > E m is that it 
can be completely described by m real -valued functions onE". (We saw 
this already in Section 4 for n = \,m = 3.) 



Sec. 7] MAPPINGS 33 

7.1 Definition Given a function F : E n — > E m , let /i,/ 2 , • • • , /» denote the 
real-valued functions on E n such that 

F(p) = (/i(p),/ 2 (p), ■■• ,/.(p)) 

for all points p in E". These functions are called the Euclidean coordinate 
junctions of F, and we write F = (fi,fe, • • • , f m )- 

The function F is differentiable provided its coordinate functions are 
differentiable in the usual sense. A differentiable function F: E n — > E TO 
is called a mapping from E n to E m . 

Note that the coordinate functions of F are the composite functions 
fi = Xi(F), where xi, • • • , x m are the coordinate functions of E m . 

Mappings may be described in many different ways. For example, 
suppose F: E —> E 3 is the mapping F = (x , yz, xy) . Thus 

F(p) = (x(p)\ y(p)z{p),x(p)y(p)) for all p. 

Now p = (pi, p2, P3), and, by definition of the coordinate functions, 

s(p) = pi> y(p) = P2» z (p) = P«- 

Hence we obtain the following pointwise formula for F: 

F (pi, P2, Pz) = {pi, PiPz, P1P2) for all p u p 2 , p 3 . 

In particular, 

F(l,-2,0) = (1,0,-2), F(-3,l,3) = (9,3,-3), 

and so on. 

In principal, one could deduce the theory of curves from the general 
theory of mappings. But curves are reasonably simple, while a mapping, 
even in the case E 2 — > E 2 , can be quite complicated. Hence we reverse this 
process and use curves, at every stage, to gain an understanding of map- 
pings. 

7.2 Definition If a : I — > E" is a curve in E n and F : E n — > E m is a mapping, 
then the composite function = F(a) : I —> E m is a curve in E m called 
the image of a under F (Fig. 1.14). 




FIG. 1.14 



34 



CALCULUS ON EUCLIDEAN SPACE 



[Chap. I 



7.3 Example Mappings. (1) Consider the mapping F: E 3 — > E 3 such 
that 

F.= (x — y, x + y, 2z). 
In pointwise terms then, 

F{VuV*>V*) = (pi - P2, Pi + p 2 , 2p 3 ) for all p u p 2 , p % . 

Only when a mapping is quite simple can one hope to get a good idea of 
its behavior by merely computing its values on some finite number of 
points. But this function is quite simple — it is a linear transformation 
from E to E . Thus by a well-known theorem of linear algebra, F is 
completely determined by its values on three (linearly independent) 
points, say the unit points 

ui= (1,0,0) u 2 = (0,1,0) u,= (0,0,1). 

(2) The mapping F: E 2 -» E 2 such that F(u,v) = (u 2 - v 2 , 2uv). (Here 
u and v are the coordinate functions of E 2 .) To analyze this mapping, we 
examine its effect on the curve a(t) = (r cos t, r sin t), where ^ t ^ 2t. 
This curve takes one counterclockwise trip around the circle of radius r 
(center at the origin.) The image curve is 

0(0 = F(a(t)) = F{r cos t, r sin t) = (r 2 cos 2 t - r 2 sin 2 £, 2r 2 cos t sin t), 

with ^ t ^ 2x. Using the trigonometric identities 

cos 2t = cos 2 t — sin 2 i, sin 2t = 2 sin t cos t, 

we find for = F(a) the formula 

0(0 = (r 2 cos2£, r 2 sin 20, 

with ^ f ^ 27r. This curve takes to#o counterclockwise trips around the 
circle of radius r 2 (center at origin) (Fig. 1.15). 

Thus the effect of F is to wrap the plane E 2 smoothly around itself twice — 
leaving the origin fixed, since ^(0,0) = (0,0). In this process, each circle 
of radius r is wrapped twice around the circle of radius r 2 . 




FIG. 1.15 



Sec. 7] 



MAPPINGS 



35 



Each time we have defined a new object in this chapter we have pro- 
ceeded to define a suitable notion of derivative of that object. For example, 
the "derivative" of a curve a is its velocity a . Using the notion of velocity 
of a curve, we shall now define the derivative F* of a mapping F: 
E n — >• E m . F* is going to be a function that assigns to each tangent vector 
v to E n at p a tangent vector F*(v) to E m at F(p). We get F*(v) by the 
following process: The tangent vector v is the initial velocity of the curve 
a (t) = p -Mv, where by Remark 3.5 we are consistently abbreviating v,, 
to simply v. Now the image of a under the mapping F is the curve £ such 
that 

0(0 = F («(*)) = F(p + <v). 

We define F*(v) to be the initial velocity 0'(O) of (Fig. 1.16). 
Summarizing this process, we obtain the following definition. 

7.4 Definition Let F: E" — ► E m be a mapping. If v is a tangent vector to 
E" at p, let F*(v) be the initial velocity of the curve t — ► F(p + tv) in E m . 
The resulting function F* (from tangent vectors of E" to tangent vectors 
of E m ) is called the derivative map F* of F. 

Note that the initial position t = of the curve t — > F(p -+- t\) is F(p). 
Thus by Definition 4.3, the point of application of its initial velocity is 
F(p). It follows from the definition, then, that F* transforms a tangent 
vector to E n at p into a tangent vector to E m at F(p). 

For example, let us compute the derivative map of the mapping 

F(u,v) = (u - v,2uv) 
in (2) of Example 7.3. For a tangent vector v at p, we have 

p + tv = (pi + tv u pi + tvt) ; 
thus 

F(p + tv) = ((pi + tv x f - (p, + tv 2 )\ 2(pi + toi)(pt + tot)). 
As t varies, this formula describes that curve in E 2 which, by definition, 



F»(v)=/3'(0), z 

F ;/l^)f>-t-*F(T>+tv) 

" /F(P)=0(O) 



(m = n = 3) 



FIG. 1.16 



36 CALCULUS ON EUCLIDEAN SPACE [Chap. I 

has initial velocity F*(v). Differentiating the coordinates above with 
respect to t (Definition 4.3), we obtain. 

F*(v) = F(p + tv)'(0) = 2(pi»i - P2V2, vvpi + P1V2) at F(p). 

7.5 Theorem Let F = (/i,/ 2 , • • • ,/») be a mapping from E n to E m . 
If v is a tangent vector to E" at p, then 

F*(v) = (v[/i], ••• ,y[fm]) atF(p). 

Thus F* (v) is determined by the derivatives v[f<] 0/ ^e coordinate functions 
of F with respect to v. 

Proof. For the sake of concreteness we take m = 3. Given v at p, we 
refer to the definition (7.4) of F* and let ft be the curve 

Pit) =F(p + tv) = (/i(p + tv),Mp + *v),/ 3 (p + tv)). 

By definition, /8'(0) = F*(v). According to Definition 4.3, to get the 
velocity vector |S'(0) we must take the derivatives at t = of the coordi- 
nate functions /i(p + tv) of 0. But (d/dt)(Ji(p + tv)) | ( = is precisely 
v[/i], where as usual the point of application p is now omitted from the 
notation. Thus 

F*(y) = Wi], v[/ 2 ], v[/ 3 ]W 

But by definition of /?, 

0(0) = F(p). I 

Fix a particular point p in E". As noted above, each tangent vector v 
to E n at p is transformed by F* into a tangent vector F* (v) to E m at F(p). 
Thus for each point p in E", the derivative map F* gives rise to a function 

F* p :T p (E n )-+T F(p) (E m ) 

which we call derivative map of F at p. Compare the corresponding situa- 
tion in elementary calculus where a differentiable function /: R — > R has 
a derivative function / : R — > R which at each point t of R gives the deriva- 
tive /(0 of /at t. 

The links between calculus and linear algebra are tighter than one might 
expect from a conventional calculus course. A most significant link is 
provided by 

7.6 Corollary Let F be a mapping from E" to E m . Then at each point 
p of E", the derivative map F* p : T p (E n ) — > T F(P )(E m ) is a linear trans- 
formation. 

Proof. If v and w are tangent vectors at p, and a and b are numbers, 
we must show that 

F*(av + 6w) = aF*(v) + 6F*(w). 



Sec. 7] MAPPINGS 37 

Using the first assertion in Theorem 3.3, this follows easily from the pre- 
ceding theorem. | 

The linearity of F* p is a generalization of the fact that the derivative 
/ (t) of /: R — ► R is the slope of the tangent line to the graph of / at t. 
Indeed for each point p, F* p is the linear transformation which best approxi- 
mates the behavior of F near p. This idea is fully developed in advanced 
calculus, where it is used to prove Theorem 7.10. 

Since F* p : T p (E n ) —> T F ( P) (E m ) is a linear transformation, it is reasona- 
ble to compute its matrix with respect to the natural bases 

tt(p), ••• , tf.(p) forr p (E n ) 

CMF(p)), ••• , U m (F(p)) for T np) (E m ). 
This matrix is called the Jacobian matrix of F at p. 
7.7 Corollary If F = (/i, • • • , f m ) is a mapping from E n to E TO , then 

F*(Uj(p)) = £ p. (p) Ui(F(p)) (1 < j < n). 
t-i ax. 

Hence the Jacobian matrix of F at p is ((dfi/dXj) (p))i^i^ m , ig>gn- 

Proof. Set v = Uj(p) in Corollary 7.6. Since the unit vector Uj(p) 
applied to /» is just (dfi/dXj) (p), we get 

'• (p ' ( >» - (I <p>> • w f <«•>)- § I <*> "'<««■»• ■ 

Standard abbreviation: 

f*(Uj) =T,¥ 1 -u i , 

i OjCj 

where Uj and dfi/dxj are evaluated at p, and V, is evaluated at F(p). 
This result shows that the derivative map of F is completely determined 
by the partial derivatives of its coordinate functions. For example, con- 
sider the second mapping in Example 7.3. Its coordinate functions are 
f = u — v and g = 2uv. Hence 



2m -2v 
2v 2w, 



Thus the Jacobian matrix of this mapping at the point p = (pi, P2) is 

'2 Pl -2p 2 \ 

2p 2 2pi 




38 CALCULUS ON EUCLIDEAN SPACE [Chap. I 

7.8 Theorem Let F: E n -* E TO be a mapping. If = F(a) is the image 
in E m of the curve a in E", then /?' = F* («'). 

This theorem asserts that F* preserves velocities of curves, since for each 
t, the velocity £ (t) of the image curve is the image, under F*, of the velocity 
a (t) of a. 

Proof. For definiteness, set m = 3. Now if F = (/ lf / 2 , / 3 ), then 

= F(a) = (/x(a),/,(a),/,(a)). 

Thus the coordinate functions of are /Si = /,(«). By Theorem 7.5, 

f.(«'(0) = («'(0 l/il, «'«) [/ 2 ], «'(0 [/,]). 

But applying Lemma 4.6, we find that 

„'«) [/.-] = «&£!>> (,) = *' (o. 

at at 

Hence 

Furthermore, this tangent vector has point of application f (a(0) = /3(0; 
hence it is precisely |3'(<). | 

Just as one uses the derivative of a function /: R — ► R to gain informa- 
tion about the function /, one can use the derivative map F* in the study 
of a mapping F. A detailed investigation of this matter belongs in advanced 
calculus; we shall give only one or two basic definitions needed in later 
work. 

7.9 Definition A mapping F: E ra — > E m is regular provided that for each 
point p of E" the derivative map F* p is one-to-one. 

Since each F* p is a linear transformation, we can apply standard results 
of linear algebra to conclude that the following conditions are equivalent: 

(1 ) F* p is one-to-one. 

(2) If F* (v„) = 0, then v p = 0. 

(3 ) The Jacobian matrix of F at p has rank n (dimension of the domain 
E n of F). 

For example, the second mapping in Example 7.3 is not regular. But 
the one-to-one condition fails at only a single point, the origin. In fact, 
the computation immediately preceding Theorem 7.8 shows that its 
Jacobian matrix has rank 2 at p ?* 0, rank at 0. 

A mapping that has an inverse mapping is called a diffeomorphism. A 
diffeomorphism is thus necessarily both one-to-one and onto, but a mapping 



Sec. 7] MAPPINGS 39 

which is one-to-one and onto need not be a diffeomorphism (Exercise 11). 
The results of this section apply equally well to mappings denned only on 
open sets of E n . In particular, we may speak of a diffeomorphism from one 
open set of E n to another. 

We state, without proof, one of the basic results of advanced calculus. 

7.1 Theorem Let F: E n — > E n be a mapping such that F* p is one-to-one 
at some point p. Then there is an open set 11 containing p such that the 
restriction of F to 1L is a diffeomorphism 11 — » V onto an open set V. 

This is called the inverse function theorem, because it asserts that the 
restricted mapping 01 — » V has an inverse mapping V — > 11. The proof is 
based on the idea that at points p + Ap very near p, F (p + Ap) is approxi- 
mately F(p) + F*(Ap). Since the tangent spaces at p and F(p) have 
the same dimension, it follows that the one-to-one linear transformation 
F* p has an inverse; hence so does F — near p. 



EXERCISES 

1. If F is the mapping F = (u 2 — v 2 , 2uv) in Example 7.3, find all points 
p such that 

(a) F(p) = (0,0). (b) F(p) = (8, -6). (c) F(p) = p. 

2. The mapping F in Exercise 1 carries the horizontal line v = 1 to the 
parabola u — > F(u,l) = (u 2 — 1, 2u). Sketch the lines u = 1 and 
v = 1, and their images under F. 

3. The image F(S) of a set S under a mapping F consists of all points 
F(p) with p in S. For F as in Ex. 1, find the image of each of the fol- 
lowing sets: 

(a) The horizontal strip S: 1 ^ v ^ 2. 

(b) The half -disc S: u 2 + v 2 ^ 1, v ^ 0. 

(c) The wedge S: —u ^ v ^ u, u ^ 0. 

In each case, show the set S and its image F(S) on a single sketch. 
(Hint: Begin by finding the image of the boundary curves of S.) 

4. (a) Show that the derivative map of the mapping (1 ) in Example 7.3 is 
given by 

F*(v P ) = (*>i - t> 2 , v i + *>2, 2v 3 )f( P ). 

(Hint: Work directly from the definition of derivative map.) 

(b) In general, if F: E n — » E TO is a linear transformation, prove that 

F*(v p ) = F(y) F(p) . 



40 CALCULUS ON EUCLIDEAN SPACE [Chap. I 

5. If F = (f h ••• ,f m ) is a mapping from E n to E m , we write 

F* = (d/i, • • • , <*/„,), 

since by Theorem 7.5, 

F*M = (dfi(\ p ), ■■■ ,df m (v p )) Hp) . 

Find F* for the mapping F = (x cos ?/, z sin y, z) from E 3 to E 3 , and 
compute F* (v p ) if 

(a) v= (2,-1,3), p = (0,0,0). 

(b) v= (2,-1,3), p= (2,t/2,»). 

6. Is the mapping in the preceding exercise regular? 

7. Let F = (/i,/ 2 ) and G = (gi,g?) be mappings from E 2 to E 2 . Compute 
the Euclidean coordinate functions of the composite function GF: 
E — > E and show that it is a mapping. 

8. In the definition (7.4) of F*(v p ), show that the straight line may be 
replaced by any curve a with initial velocity v p . 

9. Prove that a mapping F: E n — > E TO preserves directional derivatives 
in this sense: If v p is a tangent vector to E n and g is a differentiate 
function on E m , then F* (v p )\g] = \ p \g(F)]. 

10. Let F = (/i,/ 2 ) be a mapping from E 2 to E 2 . If for every point q of E 2 
the equations 

fffi = fi(Pu P*) 
have a unique solution 

fpi = giiquQ*) 

[P2 = g2(qi, q 2 ) 
prove that F is one-to-one and onto, and that F~ 1 = (^1,^2). 

11. (Continuation). In each case, show that F is one-to-one and onto, 
compute the inverse function F -1 , and decide whether F is a diffeo- 
morphism (that is, whether F -1 is differentiate). 

(a) F = (ve u , u). 

(b) F = (u 3 ,v - u). 

(c) F = (1 + 2w - 2v, 4 - 2u + 0). 

12. Let F; E n -► E m and G: E m -> E p be mappings. 

(a) Generalize the results of Exercise 7 to this case. 

(b) If a is a curve in E n )show that (£F)* («') = (?* (F* («') ). [tfmfc 
(GF)(a) = G(F( a )).] 



Sec. 8] SUMMARY 41 

(c) Deduce that (GF)* = G*F* : The derivative map of a composi- 
tion of mappings is the composition of their derivative maps. 

13. If /: R — ► R is a differentiable real-valued function on the real line 
R, prove that /* (v p ) is the tangent vector /'(p) v at the point /(p). 



8 Summary 

Starting from the familiar notion of real-valued functions, and using linear 
algebra at every stage, we have constructed a variety of mathematical 
objects. The basic notion of tangent vector led to vector fields, which 
dualized to 1 -forms— which in turn led to arbitrary differential forms. The 
notions of curve and differentiable function were generalized to that of a 
mapping F: E" — * E m . 

Then starting from the usual notion of the derivative of a real-valued 
function, we proceeded to construct appropriate differentiation operations 
for these objects: the directional derivative of a function, the exterior 
derivative of a form, the velocity of a curve, the derivative map of a 
mapping. These operations all reduced to (ordinary or partial) derivatives 
of real-valued coordinate functions, but it is noteworthy that in most cases 
the definitions of these operations did not involve coordinates. (This could 
be achieved in all cases.) Generally speaking, the differentiation operations 
all exhibited in one form or another the characteristic linear and Leibnizian 
properties of ordinary differentiation. 

Most of these concepts are probably already familiar to the reader, at 
least in special cases. But we now have careful definitions and a catalogue 
of basic properties which will enable us to begin our exploration of differ- 
ential geometry. 



CHAPTER 



II 



Frame Fields 



Roughly speaking, geometry begins with the measurement of distances 
and angles. We shall see that the geometry of Euclidean space can be 
derived from the dot product, the natural inner product on Euclidean space. 

Much of this chapter is devoted to the geometry of curves in E . We 
emphasize this topic not only because of its intrinsic importance, but also 
because the basic method used to investigate curves has proved effective 
throughout differential geometry. A curve in E is studied by assigning at 
each point a certain frame — that is, set of three orthogonal unit vectors. 
The rate of change of these vectors along the curve is then expressed in 
terms of the vectors themselves by the celebrated Frenet formulas (Theorem 
3.2). In a real sense the theory of curves inE 3 is merely a corollary of these 
fundamental formulas. 

Later on we shall use this "method of moving frames" to study a surface 
in E 3 . The general idea is to think of a surface as a kind of two-dimensional 
curve and follow the Frenet approach as closely as possible. To carry out 
this scheme we shall need the generalization (Theorem 7.2) of the Frenet 
formulas devised by E. Cartan. It was Cartan who, at the beginning of this 
century, first realized the full power of this method not only in differential 
geometry but also in a variety of related fields. 



1 Dot Product 

We begin by reviewing some basic facts about the natural inner product 
on the vector space E . 

1.1 Definition The dot product of points p = (pi, />2, P3) and q = (qi, q 2 , (ft) 
in E is the number 

p«q = piqi + ZMa + Pzq$- 
42 



Sec. 1] 



DOT PRODUCT 



43 



The dot product is an inner product, that is, it has three properties 

(1) Bilinearity: 

(ap + 6q)»r = ap»r + 6q«r 
r« (ap + 6q) = ar»p + &r»q. 

(2) Symmetry: p«q = q»p. 

(3) Positive definiteness:p*p ^ 0, andp«p = if and only if p = 0. 
(Here p, q, and r are arbitrary points of E 3 , and a and b are numbers. ) 

The norm of a point p = (pi, p 2 , Pz) is the number 

Up 11 = (p-p) 1/2 = (pi + vi + Pa 2 ) 1/2 . 

The norm is thus a real-valued function on E 3 ; it has the fundamental 
properties || p + q || ^ || p || + || q || and |[ op || = | a ||| p ||, where | a | is 
the absolute value of the number a. 

In terms of the norm we get a compact version of the usual distance 
formula in E 3 . 

1.2 Definition If p and q are points of E 3 , the Euclidean distance from 
p to q is the number 

d(p,q) = ||p - q II- 

In fact, since 

p - q = (px - q u pi - q 2 , p z - q 3 ), 

expansion of the norm gives the well-known formula (Fig. 2.1) 

d(p,q) = ((Pi ~ <fc) 2 + (P* - <? 2 ) 2 + (p, - qzf) m . 

Euclidean distance may be used to give a more precise definition of open 
sets (Chapter 1, Sectionl ). First, if p is a point of E 3 and e > is a number, 
the t-neighhorhood 9l t of p in E 3 is the set of all points q of E 3 such that 
d(p, q) < e. Then a subset of E 3 is open provided that each point of 
has an e-neighborhood which is entirely contained in 0. In short, all points 
near enough to a point of an open set are also in the set. This definition is 



X- 7 



Pi — Qs 



Pi — ?2 



FIG. 2.1 



44 



FRAME FIELDS 



[Chap. II 



valid with E replaced by E" — or indeed any set furnished with a reasonable 
distance function. 

We saw in Chapter I that for each point p of E 3 there is a canonical iso- 
morphism v — ► v p from E 3 onto the tangent space T P (E 3 ) at p. These iso- 
morphisms lie at the heart of Euclidean geometry — using them, the dot 
product on E itself may be transferred to each of its tangent spaces. 

1.3 Definition The dot product of tangent vectors v p and w p at the 
same point of E 3 is the number v p «w p = vw. 

For example, (1,0, -1)„.(3, -3,7), = 1(3) + 0(-3) + (-1)7 = -4. 
Evidently this definition provides a dot product on each tangent space 
T P (E ) with the same properties as the original dot product on E 3 . In 
particular, each tangent vector \ p toE 3 has norm (or length) || \ p || = [| v ||. 

A fundamental result of linear algebra is the Schwarz inequality 
| vw | ^ || v || || w || . This permits us to define the cosine of the angle 
# between v and w by the equation (Fig. 2.2). 



vw = 



W COS d. 



Thus the dot product of two vectors is the product of their lengths times 
the cosine of the angle between them. (The angle # is not uniquely deter- 
mined unless further restrictions are imposed, say ^ & ^ x.) 

In particular, if & = r /2, then vw = 0. Thus we shall define two vec- 
tors to be orthogonal provided their dot product is zero. A vector of length 
1 is called a unit vector. 

1.4 Definition A set d, e2 , e 3 of three mutually orthogonal unit vectors, 
tangent to E at p, is called a frame at the point p. 

Thus d, e 2 , e 3 is a frame if and only if 

ei»ei = d'd = e 3 »e 3 = 1 

ei»e 2 = ei»e 3 = e2»e 3 = 0. 




v • e 3 e 3 




v • e 2 e 2 



Sec. 1] DOT PRODUCT 45 

By the symmetry of the dot product, the second row of equations is, of 
course, the same as 

e-2*ei = e 3 «ei = e 3 »e2 = 0. 

Using index notation, all nine equations may be concisely expressed as 
e,-e,- = 8ij for 1 ^ i,j ^ 3, where 5 tJ is the Kronecker delta (0 if i 9± j, 
1 if i = j). For example, at each point p of E , the vectors t/i(p), ?7 2 (p), 
C7 3 (p) of Definition 2.4 in Chapter I constitute a frame at p. 

1.5 Theorem Let ei, 62, e 3 be a frame at a point p of E . If v is any tan- 
gent vector to E 3 at p, then (Fig. 2.3) 

v = (vei)ei + (ve2)e2 + (ve 3 )e 3 . 

Proof. First we show that the vectors e x , d, e 3 are linearly independent. 
Suppose 

22 «;«; = 0. 

Then 

= (22 a&i)'ej = 22 a,-e,-ej = 22 a * 5 *7 — a h 

where all sums are over i = 1, 2, 3. Thus 

Q>i =■ (h = a 3 = 0, 

as required. Now the tangent space T p (E ) has dimension 3, since it is 
linearly isomorphic to E 3 . Thus by a well-known theorem of linear algebra, 
the three independent vectors ei, e^, e 3 form a basis for T p (E 3 ). Hence for 
each vector v there are three (unique) numbers ci, C2, c 3 such that 

v = 22 C &i- 
But 

v*e,- = (22 c **i)' e j = 22 c &a = °h 

and thus 

v = 22 (veOc,-. | 

This result (valid in any inner-product space) is one of the great labor- 
saving devices in mathematics. For to find the coordinates of a vector v 
with respect to an arbitrary basis, one must in general solve a set of 
nonhomogeneous linear equations, a task which even in dimension 3 is 
not always entirely trivial. But the theorem shows that to find the co- 
ordinates of v with respect to a frame (that is, an orthonormal basis) it 
suffices merely to compute the three dot products vei, ve2, ve 3 . We call 
this process orthonormal expansion of v in terms of the frame ei, e%, e 3 . 



46 FRAME FIELDS [Chap. II 

In the special case of the natural frame Ui(p), £/ 2 (p), ^(p) the identity 

is an orthonormal expansion, and the dot product is denned in terms of 
these Euclidean coordinates by vw = 52 VxW%- If we use instead an arbitrary 
frame ej, 62, e 3 , then each vector v has new coordinates a* = ve» relative 
to this frame, but the dot product is still given by the same simple formula 

v«w = 52 ajbi 
since 

vw = (52 o<e.-)»(22 &***■) = S ai ^ j e *" e ' 

When applied to more complicated geometric situations, the advantage 
of using frames becomes enormous, and this is why they appear so fre- 
quently throughout this book. 

The notion of frame is very close to that of orthogonal matrix. 

1 .6 Definition Let ei, eg, e 3 be a frame at a point p of E . The 3X3 matrix 
A whose rows are the Euclidean coordinates of these three vectors is called 
the attitude matrix of the frame. 



Explicitly, if 



©1 = ( a llj a 12, 0,\z)p 
€2 = (fl21, «22, Ovijp 

e% = (031, «32, azz) P 



then 




Thus A does describe the "attitude" of the frame in E , although not its 
point of application. 

Evidently the rows of A are orthonormal, since 

52* a>ikQ>jk = ei»e, = 8ij for 1 = i,j ^ 3. 

By definition, this means that A is an orthogonal matrix. 

In terms of matrix multiplication, these equations may be written 



Sec. 1] 



DOT PRODUCT 



47 



A *A = /, where / is the 3 X 3 identity matrix, and l A is the trans-pose 
oiA: 

(an «2i «3i^ 
ai2 022 a,zz 
a U Ct23 «33/ 

It follows, by a standard theorem of linear algebra, that l AA = I, so that 
l A = A -1 , the inverse of A. 

There is another product on E 3 , closely related to the wedge product of 
1 -forms, and second in importance only to the dot product. We shall trans- 
fer it immediately to each tangent space of E . 

1.7 Definition If v and w are tangent vectors to E at the same point p, 
then the cross product of v and w is the tangent vector 

Ui(p) U t (p) U»(p) 

vXw= fi v 2 v 3 

Wi V)2 Wz 

This formal determinant is to be expanded along its first row. For ex- 
ample, if v = (1,0, -l) p and w = (2,2, -7)„ then 



v X w = 



£A(p) U t (p) Ut(p) 

1 -1 

2 2-7 



= 2*7i(p) + 5t/ 2 (p) + 2t/,(p) = (2,5,2),. 

Using familiar properties of determinants, we see that the cross product 
v X w is linear in v and in w, and satisfies the alternation rule 

v X w = — w X v 

(hence, in particular, v X v = 0). The geometric usefulness of the cross 
product is based mostly on 

1.8 Lemma The cross product v X w is orthogonal to both v and w, 
and has length such that 

|| v X w || 2 = vv ww — (vw) . 

Proof. Let v X w = ^dlhip). Then the dot product v(v X w) is 
just ^2 v^Ci. But by the definition of cross product, the Euclidean coordinates 

Ci, c 2 , Cj of v X w are such that 



v» (v X w) = 



Vi 


V2 


v 3 


Vi 


Vi 


v 3 


Wi 


w 2 


w 3 



48 FRAME FIELDS 



[Chap. II 




FIG. 2.4 



This determinant is zero, since two of its rows are the same; thus v X w 
is orthogonal to v— and, similarly, to w. 

Rather than use tricks to prove the length formula, we give a brute-force 
computation. Now 

v . vw . w _ (v-w) » = (2> 2 )(2>/) _ (j: VlWi) > 

= X) vfw* - {£ vlwi + 2 X) ViWiVjWj 
= 2J vfwf — 2^2 ViWiVjWj. 



i<] 



On the other hand, 

|| v X w|| 2 = (v X w) v (v Xwj^c, 2 

= {ViW Z — ViW 2 f + (V 3 W X — ViW 3 f + (V!W 2 — V 2 Wi) 2 

and expanding these squares gives the same result as above. | 

A more intuitive description of the length of a cross product is 

|| v X w || = || v || || w || sin 0, 

where ^ i? ^ tt is the smaller of the two angles from v to w. The direction 
of v X w on the line orthogonal to v and w is given, for practical purposes, 
by this "right-hand rule": If the fingers of the right hand point in the 
direction of the shortest rotation of v to w, then the thumb points 
in the direction ofvXw (Fig. 2.4). 

Combining the dot and cross product, we get the triple scalar product, 
which assigns to any three vectors u, v, w the number u»v X w (Exercise 4). 
Parentheses are unnecessary: u-(v X w) is the only possible meaning. 



EXERCISES 

1. Let v = (1,2,-1) and w = ( - 1 , 0, 3 ) be tangent vectors at a point of 
E 3 . Compute (a) vw. (b) v X w. 



Sec. 1] DOT PRODUCT 49 

(c) v/||v||,w/||w||. (d) ||vXw||. 
(e) the cosine of the angle between v and w. 

2. Prove that Euclidean distance has the properties 

(a) d(p,q) ^ 0; d(p,q) = if and only if p = q, 

(b) d(p,q) = d(q,p), 

(c) d(p,q) + d(q,r) ^ d(p,r), 
for any points p, q, r in E 3 . 

3. Prove that the tangent vectors 

(1, 2, 1) (-2,0,2) (1,-1,1) 

61 = V6 ' C2 = VS ' C3 = V3 

constitute a frame. Express v= (6, 1, —1) as a linear combination 
of these vectors. (Check the result by direct computation. ) 

4. Let u = (wi, t*2, Us), v = {vi,Vi,Vz), w = (u?i, 102,^3)- Prove that 



(a) u»v X w = 



Wi W2 w 3 

Vi v 2 Vz 
Wi W2 w z 



(b) u»v X w 5^ if and only if u, v, and w are linearly independent. 

(c) If any two vectors inu»v X w are reversed, the product changes 
sign. Explicitly, 

u»v X w = v»w X u = w»u X v 

= — wv X u = — vu X w = — u»w X v. 

(d) u*v X w = u X v»w. 

5. Prove that vXw^Oif and only if v and w are linearly independent, 
and show that || v X w || is the area of the parallelogram with sides v 
and w. 

6. If ei, e 2 , e 3 is a frame, show that 

ei»e 2 X e 3 = ±1. 
Deduce that any 3X3 orthogonal matrix has determinant ±1. 

7. If u is a unit vector, then the component of v in the u direction is 

vuu = II v II COS #u. 

Show that v has a unique expression v = vi + V2, where Vi«V2 = 
and vi is the component of v in the u direction. 

8. Show that the volume of the parallelopiped with sides u, v, \r is 
±u«v X w (Fig. 2.5). (Hint: Use the indicated unit vector e = v X w/ 

II v X w ||.) 



50 



FRAME FIELDS 



[Chap. II 




9. Give rigorous proofs, using e- 
neighborhoods, that each of the 
following subsets of E 3 is open: 

(a) All points p such that ||p || 
< 1. 

(b) All p such that p 3 > 0. 
(Hint: | Vi - Qi | ^ 

d(p,q)-) 

10. In each case, let S be the set of F | G . 2.5 
all points p that satisfy the 

given condition. Describe S, and decide whether it is open. 

(a) V\ + P2 2 + vi = 1- (c) pi = p 2 ^ p 3 . 

(b) Pz * 0. ( d ) Pl 2 + P2 2 < 9. 

11. If / is a differentiate function on E 3 , show that the gradient 

Vf = T,(df/dXi)U i 

(Ex. 8 of 1.6) has the following properties: 

(a) v[/] = (df ) (v) = v. (V/ ) (p) for any tangent vector at p. 

(b) The norm ||(V/)(p)|| = [Z(df/dx l )\p)'] m of (V/)(p) is the 
maximum of the directional derivatives u[/ ] for all unit vectors at p. 
Furthermore, if (V/)(p) ^ the unit vector for which the maximum 
occurs is 

(V/)(p)/||(V/)(p)||. 

The notations grad/, curl V, and div V (in the exercise referred to) 
are often replaced by V/, V X V, and V*V, respectively. 

12. Angle functions. Let/ and g be differentiable real-valued functions on 
an interval /. Suppose that f + g 2 = 1 and that # is a number such 
that/(0) = cos #0, g(0) = sin tf . If # is the function such that 

Ht) = tfo + / (fg - gf) dt, 

prove that 

/ = cos t?, g = sin &. 

(Hint: We want (/ - cos #) 2 + (g - sin i?) 2 = 0; show that 

(/cost? + gsm&)' = 0.) 

The point of this exercise is that # is a differentiable function, un- 
ambiguously defined on the whole interval /. 



Sec. 2] CURVES 51 

2 Curves 

We begin the geometric study of curves by reviewing some familiar defini- 
tions. Let a: I — »E 3 be a curve. In Chapter I, Section 4, we defined the 
velocity vector a (t) of a at t. Now we define the speed of a at t to be the 
length v(t) = || a' (Oil of the velocity vector. Thus speed is a real-valued 
function on the interval /. In terms of Euclidean coordinates a = (cm, 0-2, "3), 
we have 

Hence the speed function v of a is given by the usual formula 

In physics, the distance traveled by a moving point is determined by 
integrating its speed with respect to time. Thus we define the arc length 
of a from t = a to t = b to be the number 

f||«'(OII<a. 

Substituting the formula for || a || given above, we get the usual formula 
for arc length. 

Sometimes one is interested only in the route followed by a curve and 
not in the particular speed at which it traverses its route. One way to ig- 
nore the speed of a curve a is to reparametrize to a curve /3 which has unit 
speed || j8' || = 1. Then ft represents a "standard trip" along the route of a. 

2.1 Theorem If a is a regular curve in E 3 , then there exists a reparame- 
trization of a such that /3 has unit speed. 

Proof. Fix a number a in the domain / of a: I — > E 3 , and consider the 
arc-length function 






du. 



(The resulting reparametrization is said to be based at t = a.) Thus the 
derivative ds/dt of the function s = s(t) is the speed function v = \\ a \\ 
of a. Since a is regular, by definition a is never zero; hence ds/dt > 0. 
By a standard theorem of calculus, the function s has an inverse function 
t = t(s), whose derivative dt/ds at s = s(t) is the reciprocal of ds/dt at 
t = t(s). In particular, dt/ds > 0. 



52 FRAME FIELDS 



[Chap. II 



Now let j8 be the reparametrization 0(a) = a (<(«)) of a. We assert that 
has unit speed. In fact, by Lemma 4.5 of Chapter I, 

0'(«) = (dt/ds)(s)a(t(s)). 
Hence, by the preceding remarks, the speed of is 

llis'00 || = | (.) || «'(«(«)) || = | (s) | s («(,)) = i. i 

We shall use the notation of this proof frequently in later work. The 
unit-speed curve is sometimes said to have arc-length parametrization, 
since the arc length of from s = a to s = b (a < b) is just b — a. 

For example, consider the helix a in Example 4.2 of Chapter I. Since 
a(t) = (a cos t, a sin t, bt), the velocity a is given by the formula 

a (t) = ( — o sin t, a cos £, 6). 
Hence 

|| a'COII 1 = a(t)>a'(t) = a 2 sin 2 < + a 2 cos 2 * + 6 2 = a 2 + 6 2 . 

Thus a has constant speed c = || a \\ = (a 2 + fe 2 ) 1/2 . If we measure arc 
length from t = 0, then 



»(0 = f 

Jo 



c dt = ct. 



Hence, t(s) = s/c Substituting in the formula for a, we get the unit-speed 
reparametrization 

fi(s) = «(-) = (« cos -, a sin - , — J . 

It is easy to check directly that || j8'(s)|| = 1 for all s. 

A reparametrization a{h) of a curve a is said to be orientation-preserving 
if h > 0, orientation-reversing if h < 0. In the latter case, a and a(h) tra- 
verse their common route in opposite directions. By the conventions above, 
a unit-speed reparametrization is always orientation-preserving since 
ds/dt > for a regular curve a. 

We now define a variant of the general notion of vector field (Definition 
2.3 of Chapter I) which is adapted to the study of curves. Roughly speak- 
ing, a vector field on a curve consists of a vector at each point of the curve. 

2.2 Definition A vector field Y on a curve a : / — * E is a function that 
assigns to each number t in / a tangent vector Y(t) to E 3 at the point 
a(t). 

We have already used such vector fields, since for any curve a, its ve- 



Sec. 2] 



CURVES 



53 




FIG. 2.6 

locity a evidently satisfies this definition. Note that, unlike a , arbitrary 
vector fields on a need not be tangent to a, but may point in any direction 
(Fig. 2.6). 

The properties of vector fields on curves are analogous to those of vector 
fields on E . For example, if 7 is a vector field on a : I — > E , then for each 
t in / we can write 

7(0 = (yi(t),y2(t),y 3 (t)) a(t) = £ yS) Ui(a{t)). 

We have thus defined real-valued functions y x , y 2 , y* on / called the 
Euclidean coordinate junctions of 7. These will always be assumed to be 
differentiable. Note that the composite function t — > Ui(a(t)) is a vector 
field on a. Where it seems safe to do so, we shall often write merely £/ t 
instead of Ui(a(t)). 

The operations of addition, scalar multiplication, dot product, and cross 
product of vector fields (on the same curve) are all defined in the usual 
pointwise fashion. Thus if 

7(0 = eu, - tU 3 , Z(t) = (1 - t 2 )U 2 + tU 3 , 

and 



fit) = 



t + 1 
t 



we obtain the vector fields 

(7 + Z)(0 = * 2 t/i+ (1 - t 2 )U* 

(/7)(0 = <« + l)t/i - (*+ l)C/ 3 





t/i 


u 2 u z 


(7XZ)(0 = 


t 2 


-t 







1 - 1 2 t 



= t(l - t 2 )Ui - t 3 U 2 + t 2 (l - t 2 )U 3 



54 FRAME FIELDS [Chap. II 

and the real-valued function 

(Y-Z)(t) = -t\ 

To differentiate a vector field on a one simply differentiates its Euclidean 
coordinate functions, thus obtaining a new vector field on a. Explicitly, if 
Y = 53 y%Ui, then Y' = YL (dyi/dt)Ui. Thus, for Y as above, we get 

Y' = 2tU, - U 3 , Y" = 2U U and Y'" = 0. 

In particular, the derivative a" of the velocity a of a is called the accelera- 
tion of a. Thus if a = (ai,a2,a 3 ), the acceleration a" is the vector field 



a ~ \dt 2 ' W ~d¥/ a 



on a. By contrast with velocity, acceleration is generally not tangent to 
the curve. 

As we mentioned earlier, in whatever form it appears, differentiation 
always has suitable linearity and Leibnizian properties. In the case of 
vector fields on a curve, it is easy to prove the linearity property 

(aY + bZY = aY' + bZ' 

(a and b numbers) and the Leibnizian properties 

(/F)' = ^ Y + jY' and (Y-Z)' = Y'-Z + Y-Z'. 

If the function Y'Z is constant, the last formula shows that 

F'.Z + Y-Z' = 0. 

This observation will be used frequently in later work. In particular, if F 
has constant length || F ||, then F and F are orthogonal at each point, 
since || F || 2 = Y*Y constant implies 2F»F' = 0. 

Recall that tangent vectors are parallel if they have the same vector 
parts. We say that a vector field F on a curve is parallel provided all its 
(tangent vector) values are parallel. In this case, if the common vector 
part is (ci, c 2 , c 3 ), then 

Y(t) = (ci,c 2 ,c 3 ) aa) = X dUi for all*. 

Thus parallelism for a vector field is equivalent to the constancy of its 
Euclidean coordinate functions. 

Vanishing of derivatives is always important in calculus; here are three 
simple cases. 

2.3 Lemma (1) A curve a is constant if and only if its velocity is zero, 
a = 0. 



Sec. 2] CURVES 55 

(2) A nonconstant curve a is a straight line if and only if its acceleration 
is zero, a." = 0. 

(3) A vector field 7 on a curve is parallel if and only if its derivative 
is zero, 7 = 0. 

Proof. In each case it suffices to look at the Euclidean coordinate func- 
tions. For example, we shall prove (2). If a = (ai, as, a 3 ), then 

„ _ /(fai (fat2 d 2 aa \ 

Thus a" = if and only if each (Fai/dt 2 = 0. By elementary calculus, this 
is equivalent to the existence of constants p» and #»• such that 

cti(t) = pi + tq u for i = 1, 2, 3. 

Thus a(t) = p + tq, and a is a straight line as defined in Example 4.2 of 
Chapter I. (Note that nonconstancy implies q^O.) | 



EXERCISES 

1. For the curve a (0 = (2t, f, t 3 /3), 

(a) find the velocity, speed, and acceleration for arbitrary t, and at 

t = 1; 

(b) find the arc-length function s = s(t) (based at t = 0), and de- 
termine the arc length of a from t = — 1 to t = +1. 

2. Show that the curve a (t) = (t cos t, t sin t, t) lies on a cone in E 3 . Find 
the velocity, speed, and acceleration of a at the vertex of the cone. 

3. Show that the curve a(t) = (cosh t, sinh t, t) has arc-length function 
s (t) = y/2 sinh t, and find a unit-speed reparametrization of a. 

4. Consider the curve a(t) = (2t, t\ log t) on /: t > 0. Show that this 
curve passes through the points p = (2,1,0) andq = (4, 4, log 2), and 
find its arc length between these points. 

5. Suppose that ft and ft are unit-speed reparametrizations of the same 
curve a. Show that there is a number s such that ft(s) = ft(s + s ) 
for all s. What is the geometric significance of s ? 

6. Let Y be a vector field on the helix a(t) = (cos t, sin t, t). In each of 
the following cases, express Y in the form ^Z !/&%'■ 

(a) Y(t) is the vector from <x(t) to the origin of E 3 . 

(b) Y(t) = a'(t) -*"(t). 

(c) 7(0 has unit length and is orthogonal to both a (t) and a"{t). 

(d) 7(0 is the vector from a(t) to a(t + -rr). 



56 FRAME FIELDS [Chap. II 

7. Let F be a vector field on a curve a. If a (h) is a reparametrization of 
a, show that F(/&) is a vector field on a(h), and prove the chain rule 
Y(h)' = A' F'(/i). 

8. Let a,(i: I — »E be curves such that a (t) and /?'(£) are parallel (same 
Euclidean coordinates) for each t. Prove that a and are parallel in 
the sense that there is a point p in E 3 such that p{t) = a(t) + p for 
alH. 

9. If a is a regular curve show that 

(a) a has constant speed if and only if the acceleration a" is always 
orthogonal to a (that is, to a'). 

(b) a is a reparametrization of a straight line t — » p + tq if and only if 
a" is always tangent to a (that is, a" and a are collinear). 

10. A portion of a curve defined on a closed interval [a,b]'- a ^ t ^ b, is 
called a curve segment. A reparametrization a(h): [a,b] — »E 3 of a curve 
segment a : [c,d] — > E 3 is monotone provided either 

(a) ti ^ 0, h(a) = c, h(b) = d, or (b) ti ^ 0, h(a) = d, h(b) = c. 
Prove that monotone reparametrization does not change arc length. 

11. Prove that a straight line is the shortest distance between two points 
in E . Use the following scheme; let a : [a,b] — > E 3 be an arbitrary curve 
segment fromp = a(a) to q = a(b). Let u = (q — p)/||q — p ||. 

(a) If a is a straight-line segment from p to q, say 

a(t) = (1 - Op + *q (0 ^ t ^ 1), 

show that L(ct) = d(p, q). 

(b) From || a || ^ a'«u, deduce L(a) ^ d(p, q), where L(a) is the 
length of a and d is Euclidean distance. 

(c) Furthermore, show that if L(a) = d(p,q), then (but for para- 
metrization) a is a straight line segment. (Hint: write a = 
(a »u)u + Y, where F*u = 0.) 



3 The Frenet Formulas 

We now derive mathematical measurements of the turning and twisting of 
a curve inE . Throughout this section we deal only with unit-speed curves; 
in the next we extend the results to arbitrary regular curves. 

Let jS: / — >E be a unit-speed curve, so || j8' (s) || = 1 for each s in /. Then 
T = /3 is called the unit tangent vector field on 0. Since T has constant 
length 1, its derivative T = $" measures the way the curve is turning 
in E . We call T' the curvature vector field of 0. Differentiation of T* T = 1 
gives 2T' • T = 0, so T' is always orthogonal to T, that is, normal to /3. 



Sec. 3] THE FRENET FORMULAS 57 




The length of the curvature vector field T' gives a numerical measure- 
ment of the turning of (}. The real-valued function k such that k(s) = \\ T (s) \\ 
for all s in I is called the curvature function of 0. Thus k j^ 0, and the larger 
k is, the sharper the turning of 0. 

To carry this analysis further, we impose the restriction that k is never 
zero, so k > O.f Then the unit-vector field N = T'/k on tells the 
direction in which /3 is turning at each point. N is called the principal normal 
vector field of (Fig. 2.7). The vector field B = T X JV on is then called 
the binormal vector field of /?. 

3.1 Lemma Let /3 be a unit-speed curve in E 3 with k > 0. Then the three 
vector fields T, N, and B on are unit vector fields which are mutually 
orthogonal at each point. We call T, N, B the Frenet frame field on /S. 

Proof. By definition \\T\\ = 1. Since k = || T' || > 0, 

II n || = (iA) ii r || = i. 

We saw above that T and N are orthogonal — that is, T'N = 0. Then by 
applying Lemma 1.8 at each point, we conclude that || B \\ = 1, and B is 
orthogonal to both T and N. I 

In summary, we have T = p', N = T'/k, and B = T X N, satisfying 
T'T = N'N = B-B = 1, with all other dot products zero. 

The key to the successful study of the geometry of a curve £ is to use its 
Frenet frame field T, N, B whenever possible, instead of the natural frame 
field Ui, Ui, U 3 . For the Frenet frame field of £ is full of information about 
j9, whereas the natural frame field contains none at all. 

The first and most important use of this idea is to express the derivatives 
T', N', B' in terms of T, N, B. Since T = #', we have T' = 0" = kN. Next 
consider B' . We claim that B' is, at each point, a scalar multiple of N. 
To prove this, it suffices by ortho normal expansion to show that B »B = 
and B''T = 0. The former holds since B is a unit vector. To prove the 

t For an arbitrary unit-speed curve, this means that we must make a separate 
study of each segment on which « > 0; see Exercise 19 of Section 4. 



58 



FRAME FIELDS 



[Chap. II 



latter, differentiate B*T = 0, obtaining B' *T + B-T' = 0; then 

B'-T = -B-T' = -B-kN = 0. 

Thus we can now define the torsion function t of the curve to be the 
real-valued function on the interval I such that B' = — tN. (The minus 
sign is traditional.) By contrast with curvature, there is no restriction on 
the values of r — it may be positive, negative, or zero at various points of /. 
(Indeed the sign of t, at each point, turns out to have an interesting geo- 
metric significance.) We shall presently show that t does measure the 
torsion, or twisting, of the curve 0. 

3.2 Theorem (Frenet formulas) If /?: / — »E 3 is a unit-speed curve with 
curvature k > and torsion t, then 

T' = kN 

N' = - K T + tB. 

B' = -tN 

Proof. As we saw above, the first and third formulas are essentially just 
the definitions of curvature and torsion. To prove the second, we use 
orthonormal expansion to express N' in terms of T, N, B: 

N' = N'-T T + N'-N N + N'-B B. 

These coefficients are easily found. Differentiating N'T = 0, we get 

N'-T + N'T' = 0; hence 

N'*T = -N^T' = -N-kN = -k. 

As usual, N'*N = 0, since N is a, unit vector field. Finally, 

N'-B = -N-B' = -N-(-tN) = t. I 

3.3 Example We compute the Frenet frame T, N, B and the curva- 
ture and torsion functions of the unit-speed helix 




of \ ( s ■ s bs\ 

p(s) = o cos - , a sin - , — ) , 
\ c c c/ 

where c = (a + b 2 ) 112 and a > 0. Now 

rrf \ a' f \ ( a . s a s b\ 

T(s) = (s) = ( sin - , - cos - , - ) 

\ c c c c c) 



Hence 



T'(s) = ( — - cos -, —% sin- , 0j. 
\ c l c c 2 c / 



FIG. 2.8 



Thus 



Sec. 3] THE FRENET FORMULAS 59 

,(.) = || T'M II = « = j^ > 0. 

Since T' = nN, we get 

iV(s) = ( — cos- , — sin- , 0) 
\ c c } 

Note that regardless of what values a and b have, N always points straight 
in toward the axis of the cylinder on which /3 lies (Fig. 2.8). 
Applying the definition of cross product to B == T X N, we get 

t>< \ (b . s b s a\ 
B{s) = ( - sin - , — cos , - 1 . 
\c c c c c/ 

It remains to compute torsion. Now 

5 '( s ) = ( 1 cos " > -j sin - , J , 

and by definition, B' = — tN. Comparing the formulas for B and N, we 
conclude that 



( \ b 
t(s) = - = 



c 2 a 2 + b 2 

So the torsion of the helix is also constant. 

Note that when the parameter b is zero, the helix reduces to a circle of 
radius a. The curvature of this circle is k = l/o (so the smaller the radius, 
the larger the curvature), and the torsion is identically zero. 

This example is a very special one — in general (as the examples in the 
exercises show) neither the curvature nor the torsion functions of a curve 
need be constant. 

3.4 Remark We have emphasized all along the distinction between a 
tangent vector and a point of E 3 . However, Euclidean space has, as we 
have seen, the remarkable property that given a point p, there is a natural 
one-to-one correspondence between points (vi, #2,^3) and tangent vectors 
(vi,V2, v 3 ) p ait p. Thus one can transform points into tangent vectors (and 
vice versa) by means of this canonical isomorphism. In the next two 
sections particularly, it will often be convenient to switch quietly from one 
to the other without change of notation. Since corresponding objects have 
the same Euclidean coordinates, this switching can have no effect on scalar 
multiplication, addition, dot products, differentiation — or any other 
operation defined in terms of Euclidean coordinates. 

Thus a vector field Y = (^1,2/2,^3)0 on a curve becomes itself a curve 
(VuViyVs) in E 3 . In particular, if Y is parallel, its Euclidean coordinate 
functions are constant, so Y is identified with a single point of E . 



60 



FRAME FIELDS 



[Chap. II 



In solid geometry one describes a plane in E 3 as being composed of all 
perpendiculars to a given line at a given point. In vector language then, the 
plane through p orthogonal to q ^ consists of all points r in E 3 such that 
(r — p)»q = 0. By the remark above, we may picture q as a tangent 
vector at p as shown in Fig. 2.9. 

We can now give an informative approximation of a given curve near 
an arbitrary point on the curve. The goal is to show how curvature and 
torsion influence the shape of the curve. To derive this approximation we 
use a Taylor approximation of the curve — and express this in terms of the 
Frenet frame at the selected point. 

For simplicity, we shall consider the unit-speed curve = (ft, ft, ft) near 
the point /8(0). For s small, each coordinate ft 6s) is closely approximated 
by the initial terms of its Taylor series: 



ft(s) 



ft(0) + ^ (0) 
as 



• + &<»£ + £«»•- 



Hence 



0(s) — /3(0) + 



*'(0) + |/3"(0) +|/3'"(0). 



But jS'(O) = To, and /3"(0) = koN , where the subscript indicates evalua- 
tion at s = 0, and we assume k ^ 0. Now 

f = UN)' = ^N + kN'. 
as 

Thus by the Frenet formula for N', we get 

j8'"(0) = -ko 2 T + ^ (0) N + K ToBo. 
as 







FIG. 2.9 



0(8) ~/J(s) 

FIG. 2.10 



Sec. 3] THE FRENET FORMULAS 61 

Finally, substitute these derivatives into the approximation of 0(s) given 
above, and keep only the dominant term in each component (that is, the 
one containing the smallest power of s). The result is 

2 3 

0(s) ~ 0(0) + sT + ko I N + KOTO g #0. 

Denoting the right side by 0(s), we obtain a curve called the Frenet 
approximation of near s = 0. We emphasize that has a different Frenet 
approximation near each of its points; if is replaced by an arbitrary 
number s , then s is replaced by s — s , as usual in Taylor expansions. 

Let us now examine the Frenet approximation given above. The first 
term in the expression for is just the point 0(0). The first two terms 
give the tangent line s — ► 0(0) + sT of at 0(0)— the best linear approxi- 
mation of near 0(0). The first three terms give the parabola 

s->/8(0) + sT + ko(s 2 /2)N , 

which is the best quadratic approximation of near 0(0). Note that this 
parabola lies in the plane through 0(0) orthogonal to B , the osculating 
plane of at 0(0). This parabola has the same shape as the parabola 
y = Ko x 2 /2 in the xy plane, and is completely determined by the curvature 
ko of at s = 0. 

Finally, the torsion r , which appears in the last and smallest term of 0, 
controls the motion of orthogonal to its osculating plane at 0(0), as 
shown in Fig. 2.10. 

On the basis of this discussion, it is a reasonable guess that if a unit- 
speed curve has curvature identically zero, then it is a straight line. In fact, 
this follows immediately from (2) of Lemma 2.3, since k = || T || = || 0" ||, 
so that /t = if and only if 0" = 0. Thus curvature does measure deviation 
from straightness. 

A plane curve inE 3 is a curve that lies in a single plane of E 3 . Evidently 
a plane curve does not twist in as interesting a way as even the simple 
helix in Example 3.3. The discussion above shows that for s small the curve 
tends to stay in its osculating plane at 0(0); it is t j^ which causes 
to twist out of the osculating plane. Thus if the torsion of is identically 
zero, we may well suspect that never leaves this plane. 

3.5 Corollary Let be a unit-speed curve in E 3 with k > 0. Then 
is a plane curve if and only if t = 0. 

Proof. Suppose is a plane curve. Then by the remarks above, there 
exist points p and q such that (0(s) — p)*q = for all s. Differentiation 
yields 

0' (s) -q = 0" («) -q = for all s. 



62 




[Chap. II 



FIG. 2.11 

Thusq is always orthogonal to T = /?' and N = 0"/ K . But B is also orthog- 
onal to T and N, so, since B has unit length, B = ±q/|| q ||. Thus B f = 0, 
and by definition t = (Fig. 2.11). 

Conversely, suppose r = 0. Thus B' = 0; that is, B is parallel and may 
thus be identified (by Remark 3.4) with a point of E 3 . We assert that 
lies in the plane through (3(0) orthogonal to B. To prove this, consider the 
real-valued function 



/(*) = (0(«) ~ (()))•£ for alls. 



Then 



df 
ds 



-/ = -B = T-B = 0. 



But obviously, /(0) = 0, so /is identically zero. Thus 

(0(s) - i8(0)).5 = for alls, 

which shows that lies entirely in this plane orthogonal to the (parallel) 
binormal of /3. ■ 

We saw at the end of Example 3.3 that a circle of radius a has curvature 
1/a and torsion zero. Furthermore the formula given there for the principal 
normal shows that for a circle, N always points toward its center. This sug- 
gests how to prove the following converse. 

3.6 Lemma If is a unit-speed curve with constant curvature k > 
and torsion zero, then is part of a circle of radius \/k. 

Proof. Since r = 0, is a plane curve. What we must now show is that 
every point of is at distance \/k from some fixed point— which will thus 
be the center of the circle. Consider the curve 7 = + (1/k)N. Using the 
hypothesis on 0, and (as usual) a Frenet formula, we find 



0' +±N' = T +-{- K T) = 0. 



Sec. 3] 



THE FRENET FORMULAS 



63 



Hence the curve y is constant; that is, /3(s) + (1/k)N(s) 
has the same value, say c, for all s (see Fig. 2.12). But 
the distance from c to /3(s) is 



d(c, (S(s)) = ||c-/8(«)|| = 



- N(s) 

K 



I 




In principle, every geometric problem about curves fig. 2.12 

can be solved by means of the Frenet formulas. In simple 
cases it may be just enough to record the data of the problem in con- 
venient form, differentiate, and use the Frenet formulas. For example, 
suppose (3 is a unit-speed curve that lies entirely in the sphere 2 of ra- 
dius a centered at the origin of E 3 . To stay in the sphere, must curve; in 
fact it is a reasonable guess that the minimum possible curvature occurs 
when ^ is on a great circle of 2. Such a circle has radius a, so we con- 
jecture that a spherical curve /3 has curvature a ^ 1/a, where a is the radius 
of its sphere. 

To prove this, observe that since every point of 2 has distance a from 
the origin, we have /3«/3 = a 2 . Differentiation yields 2/3' •/? = 0, that is, 
fi*T = 0. Another differentiation gives p'*T + P'T' = 0, and by using a 
Frenet formula we get T»T + k(}*N = 0; hence 

K p.N = -1. 

By the Schwarz inequality, 

I fi.N I ^ || || || N || = a, 

and since k ^ Owe obtain the required result: 



(3-N 



Continuation of this procedure leads to a necessary and sufficient condi- 
tion (expressed in terms of curvature and torsion) for a curve to be spheri- 
cal, that is, lie on some sphere in E 3 (Exercise 10). 



EXERCISES 

1. Compute the Frenet apparatus k, t, T, N, B of the unit-speed curve 



/8(s) = (f cos s, 1 — sin s, 
find its center and radius. 



cos s). Show that this curve is a circle; 



2. Consider the curve 



0(s) 



-("■ 



+ S) 3 ' 2 (1-8) 

3 ' 3 



3/2 



'V2) 



64 



FRAME FIELDS 



[Chap. II 



defined on I: —1 < s < 1. Show that has unit speed, and compute 
its Frenet apparatus. 

3. For the helix in Example 3.3, check the Frenet formulas by direct 
substitution of the computed values of k, t, T, N, B. 

4. Prove that 

T=NXB = -BXN 
N = BXT=-TXB 
B = T X N = -N X T. 

(A formal proof uses properties of the cross product established in 
the Exercises of Section 1 — but one can recall these formulas by using 
the right-hand rule given on p. 48.) 

5. If A is the vector field tT + kB on a unit-speed curve 0, show that 
the Frenet formulas become 

T' = A X T 
N' = A XN 
B' = A XB. 

6. A unit-speed parametrization of a circle may be written 

y(s) = c + r cos s/r e x + r sin s/r e 2 , 

where e^e, = 5^. 

If is a unit-speed curve with k(0) > 0, prove that there is one 
and only one circle y which approximates £ near #(0) in the sense 
that 

7(0) = 0(0), t'(0) = /?'(0), and 7 "(0) = /?"(0). 

Show that 7 lies in the osculating plane of /3 at /3(0) and find its center 
c and radius r. The circle y is called the osculating circle and c the 



0(0) = 7(0) 





FIG. 2.13 



FIG. 2.14 



Sec. 3] THE FRENET FORMULAS 65 

center of curvature of (3 at /3(0). (The same results hold when is re- 
placed by any number s. ) 

7. If a and a reparametrization a = a(h) are both unit-speed curves, 
show that 

(a) h(s) = dts + So for some number s 

(b) f = ± T(h) 

N = N(h) H = K (h) f = r(h) 

B = ±B(h) 
where the sign (db) is the same as that in (a), and we assume k > 
(Fig. 2.14). Thus even in the orientation-reversing case, the principal 
normals N and N still point in the same direction. 

8. Curves in the plane. For a unit-speed curve /3(s) = (x(s), y(s)) inE 2 , 
the unit tangent is T = j8' = {x , y), but the unit normal N is denned 
by rotating T through +90°, so N = (-y,x). Thus T' and N are 
collinear, and the curvature of /3 is denned by the Frenet equation 
T' = kN. Prove 

(a) N' = -kT. 

(b) If <p is the slope angle f of /3, then k = <p . 

This procedure differs from that forE 3 , since k need not be positive — 
indeed its sign tells which way /3 is turning. Furthermore, N is denned 
without assuming k > 0. 

9. Let /3 be the Frenet approximation of an arbitrary unit-speed curve 
near s = 0. If, say, the B component of /3 is removed, the resulting 
curve is the orthogonal projection of /3 in the ToNq plane. It is the 
view of j8 ~ (8 one gets by looking toward /3(0) = /3(0) directly along 
the vector B . Sketch the general shape of the orthogonal projections 
of $ on each of the planes T N , T B , N B , assuming t > 0. (These 
views of P may be confirmed experimentally using a bent piece of 
wire. ) 

10. Spherical curves. Let a be a unit-speed curve with k > 0, t ^ 0. 

(a) If a lies on a sphere of center c and radius r, show that 

a — c = — pN — p aB, 

where p = 1/k and <r = 1/t. Thus r 2 = p + (p a) 2 . 

(b) Conversely, if p -f- (per) 2 has constant value r and p 7^ 0, 
show that a lies on a sphere of radius r. 

{Hint: For (b), show that the "center curve" 7 = a + pN + p <r.B — 
suggested by (a) — is constant.) 

t The existence of <p as a differentiable function with T = cos ^> C/i + sin^> ^de- 
rives from Exercise 12 of Section 1. 



66 



FRAME FIELDS 



[Chap. II 



11. Let /3, $: I — >E be unit-speed curves with nonvanishing curvature and 
torsion. If T = f , then and are parallel (Ex. 8 of II.2). If B = B, 
prove that /8 is parallel to either f} or the curve s — > —fi(s). 



4 Arbitrary-Speed Curves 

It is a simple matter to adapt the results of the previous section to the 
study of a regular curve a: I — > E 3 which does not necessarily have unit 
speed. We merely transfer to a the Frenet apparatus of a unit-speed re- 
parametrization a of a. Explicitly, if s is an arc-length function for a 
as in Theorem 2.1, then 

a(t) = a(s(t)) for alii 

or, in functional notation, a = a(s). Now if H > 0, f, t , N, and B are 
denned for a as in Section 3, we define for a the 

Curvature function: k = ic(s) 

Torsion function : t = f(s) 

Unit tangent vector field : T = f(s) 

Principal normal vector field: N = N(s) 

Binormal vector field : B = B(s) 

In general k and k are different functions, defined on different intervals. 
But they give exactly the same description of the turning of the common route 
of a and a, since at any point a(t) = a(s(t)) the numbers n(t) and k(s(0) 
are by definition the same. Similarly with the rest of the Frenet apparatus; 
since only a change of parametrization is involved, its fundamental geo- 
metric meaning is the same as before. In particular, T, N, B is again a frame 
field on a linked to the shape of a as indicated in the discussion of Frenet 
approximations. 




(— • 



FIG. 2.15 



Sec. 4] ARBITRARY-SPEED CURVES 67 

For purely theoretical work this simple transference is often all that is 
needed. Data about a converts into data about the unit-speed reparametri- 
zation a; results about a convert to results about a. For example, if a is 
a regular curve with t = 0, then by the definition above 5 has f = 0; 
by Corollary 3.5, a is a plane curve, so obviously a is, too. 

However, for explicit numerical computations — and occasionally for the 
theory as well — this transference is impractical, since it is rarely possible 
to find explicit formulas for a. (For example, try to find a unit-speed para- 
metrization for the curve a(t) = (t,f,t).) 

The Frenet formulas are valid only for unit-speed curves; they tell 
the rate of change of the frame field T, N, B with respect to arc length. 
However, the speed v of the curve is the proper correction factor in the 
general case. 

4.1 Lemma If a is a regular curve in E with k > 0, then 

T' = kvN 

N' = -kvT + tvB. 

B' = -tvN 

Proof. Let a be a unit-speed reparametrization of a. Then by definition, 
T = f(s), where s is an arc-length function for a. The chain rule as applied 
to differentiation of vector fields (Exercise 7 of Section 2) gives 

T' = f'(s) ^. 
at 

By the usual Frenet equations, f = UN . Substituting the function s in 
this equation yields 

T'(s) = k(s)N(s) = kN 

by the definition of k and N in the arbitrary -speed case. Since ds/dt is the 
speed function v of a, these two equations combine to yield T = kvN. 
The formulas for iV"' and B' are derived in the same way. | 

There is a commonly used notation for the calculus that completely 
ignores change of parametrization. For example, the same letter would 
designate both a curve a and its unit-speed parametrization a, and simi- 
larly with the Frenet apparatus of these two curves. Differences in deriva- 
tives are handled by writing, say, dT/dt for T , but dT/ds for either f 
or its reparametrization f (s). With these conventions, the proof above 
would combine the chain rule dT/dt = (dT/ds) (ds/dt) and the Frenet 
formula dT/ds = kN to give dT/dt = kvN. 

Only for a constant-speed curve is acceleration orthogonal to velocity, 
since /3 »/3 constant is equivalent to (/3 •/? ) = 2/3 «jS" = 0. In the general 



68 




[Chap. It 



FIG. 2.16 



case, we analyze velocity and acceleration by expressing them in terms of 
the Frenet frame field. 

4.2 Lemma If a is a regular curve with speed function v, then the ve- 
locity and acceleration of a are given by (Fig. 2.16) 

a = vT a" = fT + kv 2 N 
at 

Proof. Since a = a(s), where s is the arc-length function of a, we find, 
using Lemma 4.5 of Chapter I that 

a' = oc'(s) ~ = vf(s) = vT. 
at 

Then a second differentiation yields 

at at 

where we use Lemma 4.1. I 

The formula a = vT is to be expected — a and T are each tangent to 
the curve, and T has a unit length while || a \\ = v. The formula for ac- 
celeration is more interesting. By definition, a" is the rate of change of the 
velocity a , and in general both the length and the direction of a are chang- 
ing. The tangential component (dv/dt)T of a" measures the rate of change 
of the length of a (that is, of the speed of a). The normal component kv 2 N 
measures the rate of change of the direction of a. Newton's laws of motion 
show that these components may be experienced as forces. For example, 
in a car that is speeding up or slowing down on a straight road, the only 
force one feels is due to (dv/dt)T. If one takes an unbanked curve at speed 
v, the sideways force one feels is due to kv 2 N. Here k measures how sharply 
the road turns; the effect of speed is given by v 2 , so 60 miles per hour is 
four times as unsettling as 30. 



Sec. 4] ARBITRARY-SPEED CURVES 69 

We now find effectively computable expressions for the Frenet appara- 
tus. 

4.3 Theorem Let a be a regular curve in E . Then 

T =«7ll«'|| 

N = BXT k = || a X a" H/ll cl || 3 

B = a X a VII <*' X a" || r = (a X a" )•<*'"/ II «' X a" || 2 . 

Proof. Since v = \\ a \\ > 0, the formula T = a'/\\ a || is equivalent to 
a = vT. From the preceding lemma we get 

a X a" = {vT) x(f t T + kV 2 N\ 

= v ^ T X T + *v 3 r X iV = /a> 3 £ 

since T X T = 0. Taking norms we find 

|| a X a" || = || kv 3 B || = kv 3 

because || B \\ = 1, k ^ 0, and v > 0. Indeed this equation shows that for 
regular curves, \\ a X a." || > is equivalent to the usual condition k > 0. 
(Thus for k > 0, a and a" are linearly independent and determine the 
osculating plane at each point, as do T and N. ) Then 

D a' X a" a' X a" 

B = 3 



kv s ~ || a X «" || ' 

Now AT = B X T is true for any Frenet frame field (Exercise 4 of Section 
3); thus only the formula for torsion remains to be proved. To find the 
dot product (a X a" )•<*'" we express everything in terms of T, N, B. 
We already know that a X a" = wB. Thus, since = T*B = N'B, 
we need only find the B component of a". But 



«'" = (■£ T + kv 2 n\ = KV 2 N f + 



= kv\B + • • • 

where we use Lemma 4.1. Consequently (a X a")*a" = kVt, and since 
|| a X a." || = kv , we have the required formula for r. | 

The triple scalar product in this formula for t could (by Exercise 4 
of Section 1) also be written a -a" X a". But we need a X a." anyway, 
so it is usually easier to find (a X a")* a'". 

AA Example We compute the Frenet apparatus of the curve 

ait) = cm - t 3 ,3t 2 ,:u + t 3 ). 



70 



FRAME FIELDS 



[Chap. II 



The derivatives are 



Now 



a(t) = 3(1 - t\ 2t, 1 + t 2 ) 
a"(t) = 6(-«, 1,0 
«'"(0 = 6(-l,0,l). 

a(t)»a(t) = 18(1 + 2f + t 4 ), 



SO 



»«) = || a' (Oil = VT&Q. + ?). 
Applying the definition of cross product, we find 



a'(0 X a" {t) = 18 



1 - t 2 2t 1 + t 2 
-t 1 « 



= 18 (-1 + t 2 , -2t,l + t 2 ). 



Dotting this vector with itself, we get 

(18) 2 { (-1 + t 2 ) 2 + U 2 + (1 + t 2 ) 2 ) = 2(18) 2 (1 + * 2 ) 2 . 
Hence 

|| a (0 Xa"(OI| = 18V2 (1 + i 2 ). 

The expressions above for a X a" and a" yield 

(a X a"). a" = 6-18-2. 

It remains only to substitute this data into the formulas in Theorem 4.3, 
with N being computed by another cross product. The final results are 



T = 
N = 
B = 



(1 - t\ 2/, 1 + p) 
V2(l + t 2 ) 

(-2U - * 2 ,0) 
1 + t 2 

(-1 + t\ -2/,l + f) 
V2(l + t 2 ) 

1 



K = T = 



3(1 + Z 2 ) 2 



Alternatively, we could use the identity in Lemma 1.8 to compute 
|| a X a" ||, and express 

{a X a ,) •« = a • (a X a ) 

as a determinant by Exercise 4 of Section 1. 



Sec. 4] 



ARBITRARY-SPEED CURVES 



71 





FIG. 2.17 

Let us summarize the situation. We now have the Frenet apparatus for 
an arbitrary-speed curve a. This apparatus satisfies the extended Frenet 
formulas (with factor v) and may be computed by Theorem 4.3. If v = 1, 
that is, if a is a unit-speed curve, the Frenet formulas in Lemma 4.1 simplify 
slightly (to Theorem 3.2), but Theorem 4.3 may be replaced by the much 
simpler definitions in Section 3. 

Let us consider some applications of the results in this section. There 
are a number of interesting ways in which one can assign to a given curve 
/3 a new curve $ whose geometric properties illuminate some aspects of the 
behavior of /3. For example, if jS is a unit-speed curve, the curve a = T 
is the spherical image of /3. According to Remark 3.4, a is the curve 
such that each point a(s) has the same Euclidean coordinates as the unit- 
tangent vector T(s) (Fig. 2.17). Roughly speaking, (r(s) is gotten by mov- 
ing T(s) to the origin. The spherical image lies entirely on the unit sphere 
2 of E , since [|<r|| = \\ T \\ = 1, and the motion of <x represents the curving 
of P. 

For example, if £ is the helix in Example 3.3, the formula there for T 
shows that 



/N / a . s a s b\ 

r{S) = I sin - , - COS - , - J 

\ c c c c c) 



Thus the spherical image of a helix lies on the circle cut from the unit 
sphere by the plane z = b/c. 

There is no loss of generality in assuming that the original curve /? has 
unit speed, but we cannot also expect a to have unit speed. In fact, since 
a = T, we have <r = T = kN. Thus a moves always in the principal 
normal direction of /3, with speed || a || equal to the curvature k of /?. 

Next we assume k > 0, and use the Frenet formulas for to compute the 
curvature of a. Now 



as 



. K *T + ^ N + ktB. 
as 



Thus 



72 



FRAME FIELDS 



[Chap. II 



* X <;" = - K 3 N X T + ktN X B = k 2 ( k £ + rT). 
By Theorem 4.3 the curvature of the spherical image a is 



- II *' X a" 11 _ \//c 2 + r 2 



-(■+ 




2\l/2 



> 1 



and thus depends only on the ratio of torsion to curvature for the original 
curve 0. 

Here is a closely related application in which this ratio t/k turns out to 
be decisive. 

4.5 Definition A regular curve a in E 3 is a cylindrical helix provided the 
unit tangent T of a has constant angle # with some fixed unit vector u; that 
is, T(t)»-u = cos #for alU. 

This condition is not altered by reparametrization, so for theoretical 
purposes we need only deal with a cylindrical helix which has unit speed. 
So suppose is a unit-speed curve with T'-u = cos &. If we pick a reference 
point, say 0(0), on 0, then the real-valued function 

h(s) = (0(s) -0(O)). U 

tells how far 0(s) has "risen" in the u direction since leaving 0(0) (Fig. 
2.18). But 



dh 
ds 



= »u = T»n = cos # 



so is rising at a constant rate relative to arc length, and h(s) = s cos #. 
(If we shift to an arbitrary parametrization, this formula becomes 

h(t) = s(t) cos #, 

where s is the arc-length function.) 





FIG. 2.18 



FIG. 2.19 



Sec. 4] ARBITRARY-SPEED CURVES 73 

By drawing a line through each point of j8 in the u direction, we construct 
a generalized cylinder C on which ft moves in such a way as to cut each 
ruling (or "element") at constant angle #, as in Fig. 2.19. In the special 
case when this cylinder is circular, /3 is evidently a helix of the type defined 
in Example 3.3. 

It turns out to be quite easy to identify cylindrical helices. 

4.6 Theorem A regular curve a with k > is a cylindrical helix if and 
only if the ratio t/k is constant. 

Proof. It suffices to consider the case where a has unit speed. If a is a 
cylindrical helix with T*n = cos &, then 

= (T-u)' = T'*u = kN-u. 

Since k > 0, we conclude that N*u = 0. Thus for each s, u lies in the 
plane determined by T(s) and B(s). Orthonormal expansion yields 

u = cos # T + sin # B. 

As usual we differentiate and apply Frenet formulas to obtain 

= (k cos & — t sin &)N. 

Hence t sin # = k cos t?, so that t/k has constant value cot t?. 

Conversely, suppose that t/k is constant. Choose an angle # such that 
cot # = t/k. If 

U = cos # T + sin # £, 

we find 

£/' = (k cos i? - t sin d)N = 0. 

This parallel vector field U then determines (as in Remark 3.4) a unit 
vector u such that T*\i = cos t?, so a is a cylindrical helix. | 

This proof also shows how to compute the unit vector u and angle t?. 
For example, the curve a in Example 4.4 is a cylindrical helix, since k = r. 
The angle # satisfies the equation cot # = t/k = 1 ; we take & = 7r/4. 
Then cos # = sin # = 1/V2, so by the proof above, u = (1/V2) (7 1 + #)• 
The data in Example 4.4 then yield u = (0,0, 1). (There is no need to 
convert a to unit speed — that would merely reparametrize k, t, T, and B, 
with no effect on t? and u. ) 

In Exercise 10 this information about cylindrical helices is used to show 
that circular helices are characterized by constancy of curvature and tor- 
sion (see also Corollary 5.5, of Chapter III). 

Simple hypotheses on a regular curve in E 3 thus have the following 
effects (<=> means "if and only if") 



74 FRAME FIELDS [Chap. II 

k = <=> straight line 

r = <=> planar 

k constant > and t = «=> circle 

k constant > and r constant ?^ <=> circular helix 

t/k constant <=> cylindrical helix 



EXERCISES 

1. Consider the curve a: R — ► E 3 such that a(J) = (2t, t 2 , f/3). 

(a) Compute the Frenet apparatus of a: k, t, T, N, B. 

(b) Make a careful sketch of this curve for — 4 ^ t ^ 4 showing 
7\ JV, and 5 at t = 0, 2, 4. (#tnfc Begin with its projection (2t, t\ 0) 
in the xy plane.) 

(c) Find the limiting position of the Frenet frame T, N, B of a as 
t —> + oo and 2 — > — oo . 

2. Compute the Frenet apparatus of the curve a(t) = (cosh t, sinh t, t). 
Express the curvature and torsion of a as functions k(s) and t(s) 
of arc length s measured from t = 0. 

3. For the curve a(t) = (t cos t, tsint,t), 

(a) compute the Frenet apparatus at t = 0. (Evaluate a, a , ol" at 
t = before using Theorem 4.3.) 

(b) sketch this curve for — 2ir ^ t ^ 2*-, showing T, N, B a,t t = 0. 
(Hint: Ex. 2 of II.2.) 

4. For the curve a in Example 4.4, check Lemma 4.2 by direct substi- 
tution. Make a sketch, in scale, showing the vectors T(0), N(0), a (0), 
anda"(0). 

5. Prove that the curvature of a regular curve in E 3 is given by 

kV = || a" || 2 - (dv/dtf. 

6. If a is a curve with constant speed c > 0, show that 

T = a'/c k = || a" ll/c 2 

N = a"/\\ a" || _ a X a" 'a'" 

B = «' X a"A || a" || T ~ c 2 || a" || 2 

where for N, B, r, we assume a" never zero, that is, k > 0. 

7. Use the formulas in the preceding exercise to compute the Frenet 
apparatus of the helix a in Example 4.2 in Chapter I. 

8. Let a be a cylindrical helix with unit vector u, angle #, and arc-length 
function s (measured from, say, t = 0. ) The unique curve 7 such that 



Sec. 4] ARBITRARY-SPEED CURVES 75 

«(0 = 7(0 + s(t) cos #u is called the cross-section curve of the cylin- 
der on which a lies. Prove that 

(a) 7 lies in the plane through a(0) orthogonal to u. 

(b) The curvature of 7 is /c/sin 2 d, where a is the curvature of a. 
(Hint: For (b) it suffices to assume a has unit speed.) 

9. (Continuation). The following curves are cylindrical helices; for each 
find the unit vector u, angle #, and cross-section curve 7; verify con- 
dition (a) above. 

(a) The curve in Exercise 1. 

(b) The curve in Example 4.4. 

(c) The curve in Exercise 2. 

70. If is a unit-speed curve with k > and r 5* both constant, prove 
that jS is a (circular) helix. 

11. Let a be the spherical image (Section 4) of a unit-speed curve /?. 
Prove that the curvature and torsion of a are 

^ / 1 _l ( / \s (d/ ds)(r/K) 

41 + W#cn 

where « and r are the curvature and torsion of /3. 

12. (a) Prove that a curve is a cylindrical helix if and only if its spherical 

image is part of a circle. (No computations needed.) 
(b) Sketch the spherical image of the cylindrical helix in Exercise 1. 
Is it a complete circle? Find its center. 

13. If a is a curve with k > 0, then the central curve a* = a + (\/k)N 
consists of all centers of curvature of a (Ex. 6 of II.3.) For any two 
nonzero numbers a and b, let @ a b be the helix in Example 3.3. Show that 
the central curve of /8a*. is Pab, where a = — b 2 /a. Deduce that the cen- 
tral curve of && is the original helix /3„ 6 . 

14. If a(t) = (x(t), y(t)) is a regular curve in E 2 , show that its curvature 
(Ex. 8, II.3) is 

= a".J(d) = xy" - x"y 
K v* (x' 2 + y' 2 ) 3 ' 2 ' 

Here J is the operator such that 

J(tiM) = (-Mi)- 

15. For a regular curve a in E 2 , the central curve a* = a + (l/*c)iV"is 
called the evolute of a. (It is, of course, not defined at points where 

K = 0.) 

(a) Show that a* is uniquely determined by the condition that its 
tangent line at each point a*(t) is the normal line to a at a(t). 



76 FRAME FIELDS [Chap. II 

(b) Prove that 

f i 

a »J (a ) 

(notation as in Exercise 14) 

(c) Find the evolute of the cycloid 

«(0 = (t + sin tyl + cos 0, — t < t < t. 
Sketch both curves. 

16. The total curvature of a unit-speed curve a denned on I, is ji k(s) ds 
(an improper integral when the interval is infinite). If a is merely- 
regular, the definition becomes f t n(t)v(t) dt; this makes total curva- 
ture independent of the parametrization of a. Find the total curvature 
of the following curves — the first three defined on the whole real line. 

(a) The curve in Example 4.4. 

(b) The helix in Example 3.3. 

(c) The curve in Exercise 2. 

(d) The ellipse a{t) = (a cos t, b sin t). Since this is a closed curve, 
consider only a single period ^ t £ 2t. 

17. Let / > and g be arbitrary differentiable real-valued functions on 
some interval in R. Consider the curve 

«(0 = (//(*> sin *.//«) cos t,ff(t)g(t)) 

where jh denotes any function whose derivative is h. Show that the 
curvature and torsion of a are given by 



-W 1 * 



+ <7 2 + <7' 2 -1 q + g" 

T = 



+ <7 2 ) 3 / (i + g 2 + g' 2 )' 

18. Consider the general cubic curve y(t) = (at, bt 2 , ct 3 ), where abc ^ 0. 

(a) Compute 

/ = ?¥ f 9c¥ + 4bV + q 2 T /2 
T/K 26 2 |_9c 2 * 4 + 9(a 2 c 2 /6 2 )* 2 + a 2 J 

and deduce that the cubic curve 7 is a cylindrical helix if and only 
if Sac = ± 26 2 . 

(b) In the case where Sac = 2b 2 , find the unit vector u and angle #. 

19. One of the standard tricks of advanced calculus is the construction of 
an (infinitely) differentiable function / on the real line such that 
f(t) = for t ^ 0, and f(t) > for t > 0. (Also.f(0 > for t > 0.) 
If g(t) = f( — t), consider the curve 

a(t) = (t,f(t),g(t)). 



Sec. 5} 



COVARIANT DERIVATIVES 



77 



(a) Prove that the curvature of a is zero only when t = 0. 

(b) Sketch this curve for | t | small, and show some principal normals 
for t > and t < 0. 

This example shows that the condition k > cannot be avoided in a 
detailed study of the geometry of curves in E 3 , for if k is zero even at a 
single point, the geometric character of the curve can change radically at 
that point. (Note that this difficulty is not a serious one for curves in E 2 ; 
see Ex. 8 of II.3.) 



5 Cova riant Derivatives 

In Chapter I, each time we defined a new object (curve, differential form, 
mapping, . . .) we usually followed by defining an appropriate notion of 
derivative of the object. Vector fields were an exception; we have delayed 
defining their derivatives since (as later results will show) this notion 
belongs properly to the geometry of Euclidean space. 

The definition generalizes that of the derivative v[/] of a function / 
with respect to a tangent vector v at a point p (Definition 3.1 of Chapter 
I). In fact, replacing / by a vector field W, we observe that the function 
t — > W(p + tv) is a vector field on the curve t — ► p + tv. (The derivative 
of such a vector field was defined in Section 2.) Now the derivative of W 
with respect to v will be the derivative of t — > W(p + tx) at t = 0. 

5.1 Definition Let W be a vector field on E 3 and let v be a tangent vector 
to E 3 at the point p. Then the covariant derivative of W with respect to v 



WU) 




^ V W 



FIG. 2.20 



78 FRAME FIELDS [Chap. II 

is the tangent vector 

V V W = W(p + tv)'(0) 
at the point p. 

Evidently V V W measures the initial rate of change of W(p) as p moves 
in the v direction (Fig. 2.20). (The term "covariant" derives from the 
generalization of this notion discussed in Chapter VII.) 

For example, suppose W = xU x + yzU%, and v = ( — 1,0, 2) at 

P= (2,1,0). 
Then 

p + tv = (2 - t, 1, 20, 
so 

W(p + tv) = (2 - t) 2 U x + 2tU 3 , 
where strictly speaking Ui and t/ 2 are also evaluated at p + tv. Thus, 
V V W = W( V + «v)'(0) = -4C/ 1 (p) + 2C^,(p). 

5.2 Lemma If W = E «>,•£/,• is a vector field onE 3 , and v is a tangent 
vector at p, then 

V V W = £vK] Ui(p). 

Proof. We have 

W(p + tx) = X) «i(p + <v) tf.-(p + tv) 

for the restriction of W to the curve t — > p + tv. To differentiate such a 
vector field (at £ = 0), one simply differentiates its Euclidean coordinates 
(at t = 0). But by the definition of directional derivative (Definition 3.1 
of Chapter I), the derivative of w\-(p + tv) at t = is precisely v[w t -]. 
Thus 

V V W = W(p + «v)'(0) = £ vN Ui(p). | 

In short, to apply V„ to a vector field, apply v to its Euclidean coordinates. 
Thus the following linearity and Leibnizian properties of covariant deriva- 
tive follow easily from the corresponding properties (Theorem 3.3 of Chap- 
ter I ) of directional derivatives. 

5.3 Theorem Let v and w be tangent vectors to E 3 at p, and let Y and 
Z be vector fields on E 3 . Then 

(1) Vav+bwY — aV v Y + bV w Y for all numbers a and b. 

(2) V v (aY + bZ) = aV„Y + bV v Z for all numbers a and b. 



Sec. 5] COVARIANT DERIVATIVES 79 

(3) V„(/F) = v[/]F(p) +/(p) V„Fforall (differentiable) functions/. 

(4) v[Y-Z] = V„7.Z(p) + F(p).V,Z. 

Proof. For example, let us prove (4). If 

Y = 22 ViUi and Z = ^T, ZiUi, 
then 

Hence by Theorem 3.3 of Chapter I, 

v[F-Z] = v[£ y«*] = £ vfo] z t (p) + £ !fc(p) vW 

But by the preceding lemma, 

V r F = ZvIi/JC/^p) and V r Z = 2>NWp)- 

Thus the two sums displayed above are precisely V„F»Z(p) and F(p) • V„Z.| 

Using the pointwise principle (Chapter I, Section 2) once again, we can 
take the covariant derivative of a vector field W with respect to a vector 
field V, rather than a single tangent vector v. The result is the vector field 
V V W whose value at each point p is V r ( P )W. Thus V r W consists of all the 
covariant derivatives of W with respect to the vectors of V. It follows 
immediately from the lemma above that if W = ^ Willi, then 

V Y W = £ V[ Wi ]Ui. 

Coordinate computations are easy using the basic identity £/*[ /] = df/dXi. 
For example, suppose V = (y — x)U x + xyU 3 and (as in the example 
above) W = xU x + yzU z . Then 

V[x 2 ] = (y - x)UAx] = 2x(y - x) 
V[yz] = xyU 3 [yz] = xy 2 . 
Hence 

V V W = 2x(y - x)U! + xy 2 U z . 

Now the vector field V was also selected with the earlier example in mind. 
In fact, the value of V at p = (2, 1, 0) is 

V(p) = (1 - 2)C/ 1 (p) + 2*7 3 (p) = (-1,0,2), = v p , 

as before. Thus the value of the vector field V V W at this point p must 
agree with the earlier computation of V V W. And, in fact, for p = (2,1,0), 

Vr(W0(p) = 2-2(1 - 2)C/ 1 (p) + 2£/,(p) = -Wi(p) + 2U t (p). 



80 FRAME FIELDS [Chap. II 

For the covariant derivative V r W as expressed entirely in terms of vector 
fields, the properties in the preceding theorem take the following form. 

5.4 Corollary Let V, W, F, and Z be vector fields on E 3 . Then 

(1) W(aF + bZ) = aV Y Y + bV v Z, for all numbers a and b. 

(2) V f v+gwY = fV v Y + gV w Y, for all functions / and g. 

(3) V v (fY) = V[f]Y + fV v Y, for all functions/. 

(4) V[Y-Z] = VyY-Z + F-VvZ. 

We shall omit the proof, which is an exercise in the use of parentheses 
based on the (pointwise principle) definition (V F F)(p) = vV (p )F. 

Note that V F F does not behave symmetrically with respect to V and F. 
This is to be expected, since it is F that is being differentiated, while the 
role of V is merely algebraic. In particular, V /r F is/V r F, but V y (/F) is not 
/V r F: There is an extra term arising from the differentiation of / by V. 



EXERCISES 

1. Consider the tangent vector v = (1,-1,2) at the point p = (1,3, — 1). 
Compute V V W directly from the definition, where 

(a) W = x 2 U x + yU*. (b) W = xU x + x 2 U 2 - z 2 U z . 

2. Let V = — ylli + xU 3 and W = cos xUi + sin xXJi. 
Express the following covariant derivatives in terms of Ui, U%, Us: 

(a) V Y W. (c) V Y {zW). (e) V v (V r W). 

(b) V r V. (d) V W (V). (f) V v (xV - zW). 

3. If W is a vector field with constant length || W ||, prove that for any 
vector field V, the covariant derivative V V W is everywhere orthogonal 
to IF. 

4. Let X be the special vector field ^3 %iUi, where x\, x 2 , x 3 are the natural 
coordinate functions of E 3 . Prove that V r X = V for every vector field 
V. 

5. If W = 52 Willi is a vector field on E 3 , the covariant differential of W is 
defined to be VW = ^ dwi Ui. Here VW is the function on all tangent 
vectors whose value on v is 



J2dWi(v)Ui(p) = V V W. 



Compute the covariant differential of 

W = : 
and use it to find V„W, where 



W = xy Ui — x 2 z 2 Uz, 



Sec. 6] FRAME FIELDS 81 

(a) v= (1,0, -3) at p= (-1,2,-1). 

(b) v= (-1,2,-1) at p = (1,3,2). 

6. Let W be a vector field defined on a region containing a curve a. Then 
t — > W(a(t)) is a vector field on a called the restriction of W to a and 
denoted by PF«. 

(a) Prove that V a > it) W = (JF«)'(0- 

(b) Deduce that the straight line in Definition 5.1 may be replaced by 
any curve with initial velocity v. Thus the derivative Y of a vector 
field Y on a curve a is (almost) V a 'F. 

7. The bracket of two vector fields is the vector field [V, W] = V F W — V W V. 
Establish the following properties of the bracket: 

(a) [V, W][f] = VW[f] - WV[f] (here VW[f] denotes the "second 
derivative" 7[TF[/]]). 

(b) [W, V] = -[V, W\. 

(c) [U, [V, W}} + [V, [W, U]] + [W, [U, V]] = 0. 

(d) [fV, gW] = fV[g]W - gW[f]V + fg[V, W). 
{Hint: Z[f] = for all /implies Z = 0.) 



6 Frame Fields 

When the Frenet formulas were discovered (by Frenet in 1847, and inde- 
pendently by Serret in 1851), the theory of surfaces in E 3 was already a 
richly developed branch of geometry. The success of the Frenet approach 
to curves led Darboux (around 1880) to adapt this "method of moving 
frames" to the study of surfaces. Then, as we mentioned earlier, it was 
Cartan who brought the method to full generality. His essential idea was 
very simple: To each point of the object under study (a curve, a surface, 
Euclidean space itself, . . .) assign a frame; then using orthonormal ex- 
pansion express the rate of change of the frame in terms of the frame itself. 
This, of course, is just what the Frenet formulas do in the case of a curve. 

In the next three sections we shall carry out this scheme in detail for 
the Euclidean space E 3 . We shall see that geometry of curves and surfaces 
in E is not merely an analogue, but actually a corollary, of these basic 
results. Since the main application (to surface theory) comes only in 
Chapter VI, these sections may be postponed, and read as a preliminary 
to that chapter. 

By means of the pointwise principle (Chapter I, Section 2) we can 
automatically extend operations on individual tangent vectors to opera- 
tions on vector fields. For example, if V and W are vector fields on E 3 , then 



82 



FRAME FIELDS 



[Chap. II 



the dot product V'W of V and W is the (differentiable) real-valued 
function on E whose value at each point p is F(p)» W(p). The norm \\V\\ 
of V is the real- valued function on E 3 whose value at p is || F(p)||. Thus 
|| F || = (V-V) m . By contrast with V*W, the norm function || V || need 
not be differentiable at points for which V(p) = 0, since the square-root 
function is badly behaved at 0. 

At each point p of E 3 , the three tangent vectors J7i(p), U 2 (p), U 3 (p) 
constitute a frame at p. This remark is concisely expressed in terms of dot 
products of vector fields by writing UrUj = 5,-y (1 ^ i,j ^3). We used 
U h U 2 , U 3 throughout Chapter I. Now, with the dot product at our dis- 
posal, we can introduce a simple but crucial generalization. 

6.1 Definition Vector fields E t , E 2 , E 3 on E 3 constitute a frame field on 
E provided 



Ei*Ej = Si 
where 8^ is the Kronecker delta. 



(1 ^ i,j ^ 3) 



The term frame field is justified by the fact that at each point p the 
three vectors i?i(p), E 2 (p), E 3 (p) form a frame at p. We anticipated this 
in Chapter I by calling U\, 11%, U s the natural frame field on E 3 . 

6.2 Example (1) The cylindrical frame field (Fig. 2.21). Let r, &, z be 
the usual cylindrical coordinate functions on E 3 . We shall pick a unit vector 
field in the direction in which each coordinate increases (when the other 
two are held constant). For r, this is evidently 

Ei = cos d Ui + sin # U 2 , 

pointing straight out from the z axis. Then 

E 2 = - sin # Ui + cos # U 2 

points in the direction of increasing & as in Fig. 2.21. Finally the direction 




Cylindrical 

FIG. 2.21 




Spherical 

FIG. 2.22 



Sec. 6] 



FRAME FIELDS 



83 



E 3 = U 3 




FIG. 2.23 



of increase of z is, of course, straight up, so 

E 3 = U 3 . 

It is easy to check that Ei'Ej = 8a, so this is a frame field (defined on 
all of E except the z axis). We call it the cylindrical frame field on E 3 . 

(2) The spherical frame field on E 3 (Fig. 2.22). In a similar way, a frame 
field Fi, F 2 , F 3 can be derived from the spherical coordinate functions 
p, #, <p on E . As indicated in the figure, we shall measure <p up from the 
xy plane rather than (as is usually done) down from the z axis. 

Let Ei, E 2 , E 3 be the cylindrical frame field. For spherical coordinates, the 
unit vector field F 2 in the direction of increasing 
# is the same as above, so F 2 = E 2 . The unit vec- 
tor field JPi, in the direction of increasing p, points 
straight out from the origin; hence it may be ex- 
pressed as 

^1 = cos <p Ei -f- sin <p E 3 , 

(Fig. 2.23). Similarly the vector field for increas- 
ing <p is 

^3 = —sin <p Ei + cos <p E 3 . 

Thus the formulas for E u E 2 , E 3 in (1 ) yield 

Fi = cos <p(cos # Ui + sin # U 2 ) + sin <p XJ% 

F 2 = -sin # C7i + cos d- U 2 

F 3 = —sin <p(cos & Ui + sin $ U 2 ) + cos <p U 3 . 

By repeated use of the identity sin 2 + cos 2 = 1, we check that F h F 2 , F 3 
is a frame field — the spherical frame field on E 3 . (Its actual domain of defi- 
nition is E minus the z axis, as in the cylindrical case.) 

The following useful result is an immediate consequence of orthonormal 
expansion. 

6.3 Lemma Let E h E 2 , E 3 be a frame field on E 3 . 

(1 ) If V is a vector field on E 3 , then V = X) /*^« where the functions 
ft — V'Ei are called the coordinate functions of V with respect to E i} E 2 , E z . 

(2) If V = YtfiEi and W = E 9<Ei, then V-W = X)/^. In par- 
ticular, || v || = (Z//) 1/2 - 

Thus a given vector field V has a different set of coordinate functions 
with respect to each choice of a frame field E u E 2 , E 3 . The Euclidean co- 
ordinate functions (Lemma 2.5 of Chapter I), of course, come from the 
natural frame field U u U 2 , U 3 . In Chapter I, we used this natural frame 
field exclusively, but now we shall gradually shift to arbitrary frame fields. 



84 



FRAME FIELDS 



[Chap. II 



The reason is clear: In studying curves and surfaces in E 3 , we shall then be 
able to choose a frame field specifically adapted to the problem at hand. Not 
only does this simplify computations, but it gives a clearer understanding 
of geometry than if we had insisted on using the same frame field in every 
situation. 



EXERCISES 

1. If V and W are vector fields on E 3 which are linearly independent at 
each point, show that 



Ei = 



V 



V 



E» = 



W 



w\\' 



Ez = Ei X E% 



is a frame field, where W = W — W'EiEi. 

2. Express each of the following vector fields (i) in terms of the cylindrical 
frame field (with coefficients in terms of r, #, z) and (ii) in terms of the 
spherical frame field (with coefficients in terms of p, #, <p) : 

(a) Ui. (b) cos # Ui + sin # U 2 + U». 

(c) xUt + yU* + zUz. 

3. Find a frame field E\, E 2 , E 3 such that 

E l = cos x Ui + sin x cos z[/ 2 + sin x sin z £/ 3 . 

4. Toroidal frame field. Let be all of E 3 except the z axis and the circle C 
of radius R in the xy plane. The toroidal coordinate functions p, #, <p 
are defined on as suggested in Fig. 2.24, so that 




FIG. 2.24 



Sec. 7] CONNECTION FORMS 85 

x = (R + p cos ^>) cos t? 
?/ = (R + p cos p) sin # 
z = p sin <e. 

If Ei, E 2 , and E 3 are unit vector fields in the direction of increasing 
P, #, and <p, respectively, express E h E 2 , E 3 in terms of U h U 2 , U 3 and prove 
that it is a frame field. 



7 Connection Forms 

Once more we state the essential point: The power of the Frenet formulas 
stems not from the fact that they tell what the derivatives T' , N', B f are, 
but from the fact that they express these derivatives in terms of T, N, B — 
and thereby define curvature and torsion. We shall now do the same thing 
with an arbitrary frame field E u E 2 , E 3 on E 3 ; namely, express the covariant 
derivatives of these vector fields in terms of the vector fields themselves. We begin 
with the covariant derivative with respect to an arbitrary tangent vector 
v at the point p. Then 

V v Ei = CnEiip) + c 12 E 2 (p) + Ci3# 3 (p) 

V V E 2 = CuEiip) + C22# 2 (p) + C23# 3 (p) 
V V E 3 = C 3 l#l(p) + C32# 2 (p) + C33^ 3 (p) 

where by orthonormal expansion the coefficients of these equations are 

dj = V v Ei*Ej(p) for 1 S i,j ^ 3. 

These coefficients c,-y depend on the particular tangent vector v, so a 
better notation for them is 

co ti (v) = VvEi-Ejip), (1 ^ i,j ^ 3). 

Thus for each choice of i and j, on, is a real -valued function defined on all 
tangent vectors. But we have met that kind of function before. 

7.1 Lemma Let E u E 2 , E 3 be a frame field on E 3 . For each tangent vector 
v to E 3 at the point p, let 

« tV (v) = VvEi-Ejip), (1 ^ i,j ^ 3). 

Then each «,-,■ is a 1-form, and co t7 = — Wjl . These 1 -forms are called 
the connection forms of the frame field E u E 2 , E 3 . 

Proof. By definition, &>,-,■ is a real-valued function on tangent vectors, 
so to verify that w.-y is a 1-form (Definition 5.1 of Chapter I), it suffices to 



86 FRAME FIELDS [Chap. II 

check the linearity condition. But using Theorem 5.3, we get 

03ij(a\ + bw) = V av+bw Ei'Ej(p) 

= (aV v Ei + bVvE^-Ejip) 

= aV.Ei'Ej(p) + bVvErEjtp) 

= aa)ij(x) + bcoij(w). 

To prove that co»y = — a>y» we must show that w,-y(v) = — ajyi(v) for 
every tangent vector v. By definition of frame field, Ei*Ej = 8^, and since 
each Kronecker delta has constant value or 1, we have \[8i 3 ] = 0. Thus 
by the Leibnizian formula (4) of Theorem 5.3, 

= vWfEj] = VyEi-Ejip) + Ei(p).V r Ej. 

By the symmetry of the dot product, the two vectors in this last term may 
be reversed, so we have found that = &>»y(v) + w 7t (v). | 

The geometric significance of the connection forms is no mystery. The 
definition «,v(v) = V Y Ei*Ej(p) shows that co,v(v) is the initial rate at 
which E{ rotates toward Ej as p moves in the v direction. Thus the 1-forms 
Ma contain this information for all tangent vectors to E ! 

The following basic result is little more than a rephrasing of the defini- 
tion of connection forms. 

7.2 Theorem Let co;j (1 ^ i,j ^ 3) be the connection forms of a frame 
field E x , E 2 , E z on E 3 . Then for any vector field V on E 3 , 

VvEi = E <*ii(y)E h (1 S i ^ 3). 

j 

We call these the connection equations of the frame field E x , E 2 , E z . 

Proof. For fixed i, both sides of this equation are vector fields. Thus we 
must show that at each point p, 

Vv,p)^ = E ««(F(p))^y(p). 

But as we have already seen, the very definition of connection form makes 
this equation a consequence of orthonormal expansion. | 

When i = j, the skew-symmetry condition <«>„ = — wyi becomes 
thus 

COn = C022 == W33 = 0. 

Hence this condition has the effect of reducing the nine 1-forms Wjy for 



Sec. 7] CONNECTION FORMS 87 

1 ^ i,j ^ 3 to essentially only three, say co^, wu, W23. It is perhaps best to 
regard the connection forms «,-,■ as the entries of a skew -symmetric matrix 
of 1 -forms, 

(COn «12 Wu\ J C012 C0l3\ 

0>21 W22 ^23 I = I _ Wl2 0>23 

W31 W S 2 OJ33/ \ — 0)13 — U)23 

Thus in expanded form, the connection equations (Theorem 7.2) become 
V v Ei = coi2(V)Ei -f- &>i3(V)2?3 

V y # 2 = -C0 12 (F)^! + tt»(7).Bj 

Vy^J 3 = — (t)n(V)Ei — 0>23(V)^2. 

showing an obvious relation to the Frenet formulas 

T' = kN 

N' = -kT + tB 

B' = - rN. 

The absence from the Frenet formulas of terms corresponding to un{V)E 3 
and — wu (V) Ei is a consequence of the special way the Frenet frame field 
is fitted to its curve. Having gotten T(~Ei), we chose N(~E 2 ) so that 
the derivative T' would be a scalar multiple of N alone and not involve 
B(~E t ). 

Another difference between the Frenet formulas and the equations above 
stems from the fact that E 3 has three dimensions, while a curve has but one. 
The coefficients — curvature k and torsion t — in the Frenet formulas measure 
the rate of change of the frame field T, N, B only along its curve, that is, 
in the direction of T alone. But the coefficients in the connection equations 
must be able to make this measurement for E ly E i} E% with respect to ar- 
bitrary vector fields in E 3 . This is why the connection forms are 1 -forms and 
not just functions. 

These formal differences aside, a more fundamental distinction stands 
out. It is because a Frenet frame field is specially fitted to its curve that 
the Frenet formulas give information about that curve. Since the frame 
field Ei, E 2 , E 3 used above is completely arbitrary, the connection equations 
give no direct information about E 3 , but only information about the "rate 
of rotation" of that particular frame field. This is not a weakness, but a 
strength, since as indicated earlier, if we can fit a frame field to a geometric 
problem arising in E 3 , then the connection equations will give direct in- 
formation about that problem. Thus these equations play a fundamental 
role in all the differential geometry of E 3 . In particular, the Frenet formulas 



88 FRAME FIELDS [Chap. II 

can be deduced from them (Exercise 8). For the sake of motivation, however, 
we have preferred to deal with the simpler Frenet case first. 

Given an arbitrary frame field E x , E 2 , E 3 on E 3 , it is fairly easy to find an 
explicit formula for its connection forms. First use orthonormal expansion 
to express the vector fields Ei, E 2 , E 3 in terms of the natural frame field 
Ur, U 2 , C/ 3 onE 3 : 

Ei = a n Ui + a 12 U 2 + a n U 3 
E 2 = (hiUi + ChiUi + ckaUz 
Ez = CL31U1 -+- a®,U 2 + azsUz. 

Here each a t -y = Ei* U, is a real-valued function on E 3 . The matrix 

(«n «i2 aiz^ 
«2i an a 2 z 
azi a 3 2 033/ 

with these functions as entries is called the attitude matrix of the frame 
field Ei, E 2 , E s . In fact, at each point p, the numerical matrix 

A(p) = (o,-,-(p)) 

is exactly the attitude matrix, of the frame 2?i(p), E 2 (p), E 3 (p) as in Defi- 
nition 1.6. Since attitude matrices are orthogonal, the transpose l A of A 
is equal to its inverse A -1 . 

Define the differential of A = {an) to be dA = (da t -y), so dA is a matrix 
whose entries are 1 -forms. We can now give a simple expression for the 
connection forms in terms of the attitude matrix. 

7.3 Theorem If A = (a t y) is the attitude matrix and w = (co t y) the 
matrix of connection forms of a frame field E u E 2 , E z , then 

03 = dA l A (matrix multiplication), 

or, equivalently 

«»y = 2 a ik da ik for 1 ^ i,j ^ 3. 

k 

Proof. If v is a tangent vector at p, then by definition, 

« t7 (v) = VvEi-Ejip). 
Since A is the attitude matrix, 

Ei = X) <**U k , 
and thus, by Lemma 5.2, 



Sec. 7] CONNECTION FORMS 89 

VvEi = £v[o ft ] U k (p). 
The dot product of this vector with 

Ej(p) = 2> ifc (p) U k (p) 
is then 

k 
But by the definition of differential, 

v[a lfc ] = da ik {\); 
hence 

w »i(v) = ^2a jk (p) da ik (\) = (^ a jfc da tfc ) (v). 

Since this equation holds for all tangent vectors, the two 1 -forms w»_,- and 
53 o,jk da ik are equal. It is easy to get the neater matrix formula. In fact, 
the transpose l A has entries l a k j = aj k , so 

«»•; = 2 da ik l a k j for 1 ^ ij ^ 3, 

which in terms of matrix multiplication is just co = dA l A. | 

Using this result, let us compute the connection forms of the cylindrical 
frame field in Example 6.2. From the definition, we read off the attitude 
matrix 

cos & sin # 0^ 

A = | — sin# cost? 

L 



Thus 



Since 



'-sintfdtf costfdtf 0^ 
dA = | — costfdtf — sintfdtf 
Oy 

''cost? — sint? 0^ 
*A = ( sin tf cos # ) , 
1, 



90 FRAME FIELDS [Chap. II 

we easily compute 

d& 0^ 

a = dA l A = | -d& 

0/ 

Thus o>i2 = d& and all the other connection forms (except, of course, 
co 2 i = — co i2 ) are zero. Then the connection equations (Theorem 7.2) of the 
cylindrical frame field become 

VyEi = d&(V)E 2 = V[#]E 2 
V V E 2 = -d&(V)E l = -V[&\Ei 
V V E 3 = 

for all vector fields V. 

These equations have obvious geometric significance. The third equa- 
tion says that the vector field E 3 is parallel. We knew this already, since in 
the cylindrical frame field, E 3 is just U 3 . The first two equations say that 
the covariant derivatives of E x and E% with respect to an arbitrary vector 
field V depend only on the rate of change of the angle & in the direction of 
V. From the way the function # is defined, it is clear that V[&] = when- 
ever V is, at each point, tangent to the plane through the z axis. Thus for 
a vector field of this type, the connection equations above predict that 
V v Ei = V v Ei = 0. From Fig. 2.21, it is clear that E 1 and E 2 do stay parallel 
on any plane through the z axis. 



EXERCISES 

1. For any function /, show that the vector fields 

E, = (sin fU, + U* - cosfU 3 )/V2 
E 2 = (sin/C/x - U* - cosfU 3 )/V2 
E z = cos fU\-\- sin / U 3 
form a frame field, and find its connection forms. 

2. Find the connection forms of the natural frame field Ui, U 2 , U 3 . 

3. For any function /, show that 



cos / 


cos / sin / 


sin/ 


sin / cos / 


• 2 -e 

sm / 


-cos/ 


sin/ 


cos/ 






is the attitude matrix of a frame field, and compute its connection 
forms. 



Sec. 8] THE STRUCTURAL EQUATIONS 91 

4. Prove that the connection forms of the spherical frame field are 

wi2 = cos <p d&, coi3 = d<p, C023 = sin <p d&. 

5. If Ei, E 2 , E s is a frame field and W = ^ fiE i? prove the covariant de- 
rivative formula: 

v r w = Zmfj] + I/wTO. 

J i 

6. Let Ei, Ei, E z be the cylindrical frame field. If V is a vector field such 
that V[&] = 1, compute vV(r cos # #1 + r sin # ^/ 3 ). 

7. If Fi, F 2 , Fz is the spherical frame field, 

(a) Prove that Fi[p] = 1 and FM] = FM = 0. 

(b) Compute V, x (cos p F 2 + sin p F 3 ). 

8. Let j8 be a unit-speed curve in E 3 with k > 0, and suppose that i?i, E 2 , E 3 
is a frame field on E 3 such that the restriction of these vector fields to 
|8 gives the Frenet-frame field T, N, B of 0. Prove that 

Mir) = k, W i,(r) = o, co 23 (r) = r. 

Then deduce the Frenet formulas from the connection equations. {Hint: 
Ex. 6 of II.5.) 



8 The Structural Equations 

We have seen that 1 -forms — the connection forms — give the simplest 
description of the rate of rotation of a frame field. Furthermore, the frame 
field itself can be described in terms of 1 -forms. 

8.1 Definition If E x , E 2 , E 3 is a frame field on E 3 , then the dual l-forms 
0i, 02, 03 of the frame field are the l-forms such that 

Bi(y) = vEi(p) 
for each tangent vector v to E at p. 

Note that 0» is linear on the tangent vectors at each point; hence it is a 
1-form. (Readers familiar with the notion of dual vector spaces will recog- 
nize that at each point, 0i, 2 , 03 gives the dual basis of E u E 2 , E 3 .) 

In the case of the natural frame field Ui, U 2 , Uz, the dual forms are just 
dxi, dx 2 , dx 3 . In fact, from Example 5.3 of Chapter I we get 

dxi{\) = Vi = v[/,-(p) 

for each tangent vector v; hence dxi = t . 

Using dual forms, the orthonormal expansion formula in Lemma 6.3 



92 FRAME FIELDS [Chap. II 

may be written V = E 9i{V)Ei. In the characteristic fashion of duality, 
this formula becomes the following lemma. 

8.2 Lemma Let 0i, 2 , 03 be the dual 1 -forms of a frame field E u E 2 , E 3 . 
Then any 1-form <j> on E has a unique expression 

Proof. Two 1 -forms are the same if they have the same value on any 
vector field V. But 

(2>w*i)(J0 = 2>CE<)fc<n 

These functions <f>(Ei) are the only possible coordinate functions for <j> in 
terms of 6 U B 2 , 03, since if <f> = E fid iy then 

<f>(Ej) = E fMEj) = E /Ay = /y- I 

Thus <f> is expressed in terms of dual forms of Ei, E 2 , E 3 by evaluating it 
on Ei, E2, E 3 . This useful fact is the generalization to arbitrary -frame fields 
of Lemma 5.4 of Chapter I. 

We compared a frame field E u E 2 , E 3 to the natural frame field by means 
of its attitude matrix A = {an), for which 

Ei = E *aUi (l£i£ 3). 

The dual formulation is just 

6i — E °*'i dxj 

with the same coefficients. In fact, by Lemma 8.2 (or rather its special case, 
Lemma 5.4 of Chapter I), we have 

Oi = E 9i{Uj) dxj. 

But 

di(Uj) = Ei-Uj = (E aikU k )'Uj = an. 

These formulas for Ei and 0; show plainly that 0i, 2 , 3 is merely the dual 
description of the frame field E u E 2 , E 3 . 

In calculus, when a new function appears on the scene, it is natural to 
ask what its derivative is. Similarly with 1-forms — having associated with 
each frame field its dual forms and connection forms — it is reasonable to 
ask what their exterior derivatives are. The answer is given by two neat 
sets of equations discovered by Cartan. 

8.3 Theorem (Cartan structural equations) Let E x , E 2 , E 3 be a frame 
field on E 3 with dual forms X , 2 , 03 and connection forms 0,,, (1 ^ i,j ^ 3). 
The exterior derivatives of these forms satisfy 



Sec. 8] THE STRUCTURAL EQUATIONS 93 

(1 ) the first structural equations: 

ddi = 2 «« A e > (1 ^ * ^ 3); 

(2) the second structural equations: 

dun = 22 «»* A w *y (1 = *J = 3). 

it 

Because 0,- is the dual of E i7 the first structural equations may be easily 
recognized as the dual of the connection equations. Only on the basis of 
later experience will we discover that the second structural equation shows 
that E 3 is flat — roughly speaking, in the same sense that the plane E 2 is flat. 

Proof. We have seen that 

hence 

ddi = 22 dttij a dxj. 

Since the attitude matrix A = (an) is orthogonal, the expression in Theo- 
rem 7.3 for wn in terms of dan may be solved for dan by the usual for- 
malism of linear algebra to give 

dan = 2-j w utQ>kj' 

k 

Thus 

ddi = 22{ (22 uika k j) A dxj] 

i k 

= 22 {vat A /2 aki dxj\ 

k i 

— 22 w * A ^fc 

k 

which is the first structural equation. 

We could give a similar index proof for the second structural equation, 
but a liberal use of matrix notation will give a better idea of what is really 
going on. To apply the exterior derivative d to a matrix of functions or 
1 -forms, we apply it to each entry. The matrix formula go = dA l A in 
Theorem 7.3, for example, means 

w,-y = 2^ daikajk. 
But 

dwij = — /] danc A dajk. 

(Note the minus sign!) Hence in matrix notation, where we shall suppress 



94 FRAME FIELDS [Chap. II 

the wedge, 

doo = -dA l (dA). 
Multiplying both sides of co = dA l A by A yields 

dA =uA, 
since A = A~\ Then the rule for transpose of a matrix product gives 

l (dA) = *M) = l A l oo. 
By substituting in the equation above for do: we find 



do: = — coA A co = — 



co co = coco 



since co is skew -symmetric. But this is just the second structural equation 
dun = ^2 ooik A oikj in matrix notation. | 

8.4 Example Structural equations for the spherical frame field (Ex- 
ample 6.2). The dual forms and connection forms are 

0i = dp coi2 = cos <p d& 

2 = p cos (p d& coi3 = dtp 

3 = p dip co 2 3 = sin <p d&. 

Let us check, say, the first structural equation 

d8 3 = 2_, &3j A 8j = CO31 A Q 1 -\- tojg a # 2 . 

Using the skew -symmetry co t y = — co.,-; and the general properties of forms 
developed in Chapter I, we get 

C031 A 81 = —d<p a dp = dp a d<p 

CO32 a 2 = ( — sin cp d#) a (pcos<pd#) = 

(the latter since ci# a d# = 0). The sum of these terms is, correctly, 

d8z = d(p d(p) = dp a dip. 

Second structural equations involve only one wedge product. For ex- 
ample, since con = co 2 2 = 0, 

dun = 2^ "i* A ^ = C013 a W32 . 

In this case, 

oJn A C032 = dtp a (— sin tp d&) = — sin tp dip a d& 

which is the same as 

cicoi2 = d(cosipd&) = d (cos ip) a dd 3= — sin <p d<p a rf#. 



Sec. 8] THE STRUCTURAL EQUATIONS 95 

Fi 




p COS <p flW 



FIG. 2.25 



To derive the expressions given above for the dual 1 -forms, first compute 
dxi, dx2, dx 3 by differentiating the well-known equations 

Xi = p cos <p cos & 
x 2 = p cos <p sin t? 
x% = p sin <p. 

Then substitute in the formula t = ^ a^ dx h where A = (a t y) is the 
attitude matrix to be found in Example 6.2. 

We shall later find a more efficient computational technique related to 
the following reliable nonsense from elementary calculus: If at each point 
the spherical coordinates p, #, <p are altered by dp, d&, d<p, then the sides of 
the resulting infinitesimal box are dp, p cos <p d&, p d<p (Fig. 2.25). But these 
are precisely the formulas for the dual forms 0i, 02, 3 . 

As we mentioned earlier, our main use of the Cartan structural equations 
is in studying the geometry of surfaces (Chapters VI and VII). A wider 
variety of applications is given by Flanders [1]. 



EXERCISES 

1. For a 1-form <f> = ^2 fiQi, prove that 

d<t> = E {dfj + L/wy) A 8j- 

i 

(Compare Ex. 5 of II.7.) 

2. For the toroidal frame field in Ex. 4 of II.6, show that 



96 FRAME FIELDS [Chap. II 

#i = dp 0012 = cos <p d& 

2 — (R + p cos <p)d& a>i3 = d<p 

03 = p d<p C023 = sin <p d&. 

(Hint: Find 0* by the scheme described at the end of II.8. No computa- 
tions are necessary to find caul) 

3. Check the first structural equations in the case of the toroidal frame 
field. 

4. For the cylindrical frame field E h E 2 , E 3 : 

(a) Prove that 0i = dr, 2 = r d&, 3 = dz by evaluating on U u U 2 , U 3 . 

(b) Deduce that EM = 1, E 2 [&] = 1/r, E 3 [z] = 1 and that the other 
six possibilities Ei[&\, • • • are all zero. 

(c) For a function /(r, t?, 0) expressed in terms of cylindrical coordi- 
nates, show that 

m -l Ew-lg Bin- 1 

5. Frame fields on E 2 . For a frame field E u E 2 on the plane E 2 : 

(a) Find the connection equations. 

(b) If 

Ei = cos <p Ui + sin <p U 2 

E 2 = — sin (pUi -\- cos <p U 2 

where ^ is an arbitrary function, express the dual 1-forms 0i, 2 
and the connection form co i2 in terms of <p. 

(c ) Prove the structural equations for this case. 



9 Summary 

We have accomplished the aims set at the beginning of this chapter. The 
idea of a moving frame has been expressed rigorously as & frame field — either 
on a curve in E , or on an open set of E itself. In the case of a curve, we 
used only the Frenet frame field T, N, B of the curve. Expressing the 
derivatives of these vector fields in terms of the vector fields themselves, 
we discovered the curvature and torsion of the curve. It is already clear 
that curvature and torsion tell a lot about the geometry of a curve; we 
shall find in Chapter III that they tell everything. In the case of an open 
set of E 3 , we dealt with an arbitrary frame field E u E 2 , E 3 . Cartan's gener- 
alization (Theorem 7.2) of the Frenet formulas followed the same pattern 
of expressing the (covariant) derivatives of these vector fields in terms of 



Sec. 9] SUMMARY 97 

the vector fields themselves. Omitting the vector field V from the notation 
in Theorem 7.2 we have 

Cartan Frenet 

V#i = CO12.EJ2 + COi3#3 T ' = KN 

VEi = -wiJ^i + W23-E3 N' = -kT + t£ 

Cartan's equations are not conspicuously more complicated than Fre- 
net 's because the notion of 1-form is available for the coefficients &>,•/, the 
connection forms. 



CHAPTER III . 

Euclidean Geometry 



We recall some familiar features of plane geometry. First of all, two tri- 
angles are congruent if there is a rigid motion of the plane which carries 
one triangle exactly onto the other. Corresponding angles of congruent 
triangles are equal, corresponding sides have the same length, the areas 
enclosed are equal, and so on. Indeed, any geometric property of a given 
triangle is automatically shared by every congruent triangle. Conversely, 
there are a number of simple ways in which one can decide if two given 
triangles are congruent— for example, if for each the same three numbers 
occur as lengths of sides. 

In this chapter we shall investigate the rigid motions (isometries) of 
Euclidean space, and see how these remarks about triangles can be extended 
to other geometric objects. 



1 Isometries of E 3 

An isometry, or rigid motion, of Euclidean space is a special type of map- 
ping which preserves the Euclidean distance between points (Defini- 
tion 1.2, Chapter II). 

1.1 Definition An isometry of E 3 is a mapping F: E 3 — ► E 3 such that 

d(F(p),F(q)) =d(p,q) 
for all points p, q in E 3 . 

1.2 Example 

(1) Translafions. Fix a point a in E 3 and let T be the mapping that adds 
a to every point of E 3 . Thus T^p) = p + a for all points p. T is called 
translation by a. It is easy to see that T is an isometry, since 

98 



Sec. 1] 



ISOMETRIES OF E 3 



99 




FIG. 3.1 



d(T(p),T(q)) = tf(p + a,q + a) 

= ||(p + a) - (q + a)|| 

= Up -qll = d(p,q). 

(2) Rotation around a coordinate axis. A rotation 
of the xy plane through an angle # carries the 
point (pi, p 2 ) to the point (51,92) with coordi- 
nates (Fig. 3.1) 

Qi = Pi cos & — P2 sin # 

#2 = Pi sin # + P2 cos #. 

Thus a rotation C of E around the 2 axis (through an angle t?) has the 
formula 

C(p) = C (pi, p 2 , p 3 ) = (pi cos # — p 2 sin #, pi sin # + p 2 cos &, p 3 ) 

for all points p. Evidently, C is a linear transformation, hence, in particular, 
a mapping. A straightforward computation shows that C preserves Euclid- 
ean distance, so C is an isometry. 

Recall that if F and G are mappings of E 3 , the composite function GF 
is a mapping of E 3 obtained by applying first F, then G. 

1.3 Lemma If F and G are isometries of E 3 , then the composite mapping 
GF is also an isometry of E 3 . 

Proof. Since G is an isometry, the distance from G(F(p)) to (?(F(q)) 
is d(F(p), F(<i)). But since F is an isometry, this distance equals d(p, q). 
Thus GF preserves distance, hence is an isometry. | 

In short, a composition of isometries is again an isometry. 

We also recall that if F: E 3 — » E 3 is both one-to-one and onto, then F 
has a unique inverse function F _1 : E 3 — > E 3 , which sends each point F(p) 
back to p. The relationship between F and F~* is best described by the 
formulas 

FF~ l = I, F' X F = I 

where / is the identity mapping of E 3 , that is, the mapping such that 

^(p) = P f° r an P- 

The translations of E (as defined in Example 1.2) are the simplest 
type of isometry. 



1.4 Lemma 
translation. 



(1) If aS and T are translations, then ST = TS is also a 



100 EUCLIDEAN GEOMETRY [Chap. Ill 

(2) If T is translation by a, then T has an inverse T~\ which is trans- 
lation by — a. 

(3) Given any two points p and q of E 3 , there exists a unique transla- 
tion T such that T(p) = q. 

Proof. To prove (3), for example, note that translation by q - p cer- 
tainly carries p to q. This is the only possibility, since if T is translation 
by a and T(p) = q, thenp + a = q; hence a = q - p. | 

A useful special case of (3) is that if T is a translation such that for some 
one point T(p) = p, then T = I. 

The rotation in Example 1.2 is an example of an orthogonal transforma- 
tion of E , that is, a linear transformation C: E 3 — > E 3 which preserves 
dot products in the sense that 

C(p)»C(q) = p»q forallp,q. 

1.5 Lemma If C: E 3 — > E 3 is an orthogonal transformation, then C is 
an isometry of E 3 . 

Proof. First we show that C preserves norms. By definition ||p || 2 = p.p; 
hence 

IIC(p)|| 2 = C(p).c( P ) = P .p= ||p || 2 . 

Thus || C(p)|| = ||p || for all points p. Since C is linear, it follows easily 
that C is an isometry: 

d(C(p), C(q)) = || C(p) - C(q)|| = || C(p - q)|| = ||p - q.|| 

= d(p, q) for all p, q. | 

Our goal now is Theorem 1.7, which asserts that every isometry can 
be expressed as an orthogonal transformation followed by a translation. 
The main part of the proof is the following converse of Lemma 1.5. 

1.6 Lemma If F is an isometry of E 3 such that ^(0) = 0, then F is 
an orthogonal transformation. 

Proof. First we show that F preserves dot products; then we show that 
F is a linear transformation. Note that by definition of Euclidean distance, 
the norm || p || of a point p is just the Euclidean distance d(0, p) from the 
origin top. By hypothesis, F preserves Euclidean distance, and F(0) = 0; 
hence 

|| F(p)\\ = d(0, F(p)) = d(F(0), F(p)) = d(0,p) = ||p ||. 

Thus F preserves norms. Now by a standard trick ("polarization") we 
shall deduce that it also preserves dot products. Since F is an isometry, 



Sec. 1] ISOMETRIES OF E 3 101 

d{F(p),F(q)) =rf(p,q) 

for any pair of points. Hence 

\\F(p)-F( q )\\ = ||p -q |l, 

which, by definition of norm, implies 

(F(p)-F(q)).(F(p) -F(q)) = (p-q)-(p-q). 

Hence 

|| F(p)f - 2F<p).F(q) + || F(q)|| 2 = ||p II 2 - 2p.q + || q || 2 . 

The norm terms here cancel, since F preserves norms, and we find 

F(p)-F(q) =p.q, 

as required. 

It remains to prove that F is linear. Let ui,u 2 ,u 3 be the unit points 
(1, 0, 0), (0, 1, 0), (0, 0, 1), respectively. Then we have the identity 

p = (pi,p2,pz) = 2 Pi^i- 

Also, the points ui,U2,U3 are orthonormal; that is, u,-uy = bij. 

We know that F preserves dot products, so F{u x ), F(u 2 ), F(u 3 ) must 
also be orthonormal. Thus orthonormal expansion gives 

F(p) = ZF^-FMFiui). 
But 

F(p)-F(ui) =p*iii = p h 
so 

F(p) = T,PiF( Ui ). 
Using this identity, it is a simple matter to check the linearity condition 
F(ap + 6q) = aF(p) + bF(q). | 

We now give a concrete description of what an arbitrary isometry is like. 

1.7 Theorem If F is an isometry of E 3 , then there exist a unique trans- 
lation T and a unique orthogonal transformation C such that 

F = TC. 

Proof. Let T be translation by F(0). We saw in Lemma 1.4 that T~ l 
is translation by — F(0). But T~ l F is an isometry, by Lemma 1.3, and 
furthermore, 

(T -1 /0(0) = T~\F(0)) = F(0) - F(0) = 0. 



102 EUCLIDEAN GEOMETRY [Chop. Ill 

Thus by Lemma 1.6, T~*F is an orthogonal transformation, say T~*F = C. 
Applying T on the left, we get F = TC. 

To prove the required uniqueness, we suppose that F can also be ex- 
pressed as TC, where f is a translation and C an orthogonal transformation. 
We must prove f = T and C = C. Now TC = TC; hence C = T~ l TC. 
Since C and C are linear transformation, they of course send the origin to 
itself. It follows thatjT _1 f)(0) = 0. But since T~ l f is a translation, 
we conclude that T~'T = I; hence T = T. Then the equation TC = TC 
becomes TC = TC. Applying T~\ we get C = C. | 

Thus every isometry of E 3 can be uniquely described as an orthogonal trans- 
formation followed by a translation. When F = TC as in Theorem 1.7, we 
call C the orthogonal part of F, and T the translation part of F. Note that 
CT is generally not the same as TC (Exercise 1). 

The decomposition theorem above is the decisive fact about isometries 
of E 3 (and its proof holds for E" as well). For example, we shall now find 
explicit formulas for an arbitrary isometry F = TC. If (c,-y) is the ma- 
trix of the linear transformation C, we have the explicit formula 

C(Pii P2, ps) = (S CijPj, H c*,Vh 2 c 3j pj) 

for all points p = (pi,Pi,pz). We are using the column-vector conventions, 
under which q = C(p) means 

^A /Cll Ci 2 C13 

?2 I = I Cn C 22 C 23 

\W \c 3 i C S2 c M . 

Since C is an orthogonal linear transformation, it is easy to show that its 
matrix (dj) is orthogonal in the sense: inverse equals transpose. 

Returning to the decomposition F = TC, suppose that T is translation 
by a = (a u a 2 , a 3 ). Then 

F(p) = TC(p) = a + C(p). 

Using the above formula for C(p), we get 

F(p) = F(p h p 2 ,p 3 ) = (oi + £ ciip h 02 + X) c 2 y?>i, a 3 + 2 <&#,■). 

Alternatively, using the column-vector conventions, q = F(q) means 

f Qi\ /oA / c " C 12 ci 3 \ /p^ 
g 2 J = I a 2 J + I c 2 i C22 c 2 3 I I P2 

\qJ W/ Vsi c 32 C33/ \p 3 y 




Sec. 1] ISOMETRIES OF E 3 103 

EXERCISES 

Throughout these exercises, A, B, and C denote orthogonal transforma- 
tions (or their matrices), and T a is translation by a. 

1. Prove that CT a = T CW C 

2. Given isometries F = T a A and G = T b B, find the translation and 
orthogonal part of FG and GF. 

3. Show that an isometry F = T a C has an inverse mapping F~ x , which is 
also an isometry. Find the translation and orthogonal parts of F~ l . 

4. If 

C = [ i I -f ] and ( P = , (3 ' *' ~ 6) ' 



1 2 
3 3 



q= (1,0,3) 




show that C is orthogonal; then compute C(p) and C(q), and. check 
thatC(p).C(q) =p.q. 

5. Let F = T a C, where a = (1,3,-1) and 

/1/V2 
C = 

\1/V2 

If p = (2, —2, 8), find the coordinates of the point q for which 
(a)q = F(p). (b)q = ^(p). (c) q = (CT )(p). 

6. In each case decide whether F is an isometry of E 3 . If so, find its trans- 
lation and orthogonal parts. 

(a) F(j>) = -p. 

(b) ^(p) = p»a a, where || a || = 1. 

(c) F(p) = (vz- 1,P2-2, V1 - 3). 
(d)F(p) = (vuP2,l). 

A group G is a set furnished with an operation that assigns to each 
pair g lf g 2 of elements of G an element <7i0 2 , subject to these rules: (1) 
associative law: (gig^gs = ?ite), (2) there is a unique identity 
element e such that eg = ge — g for all g in G, and (3) inverses: For 
each g in G there is an element gr _1 in G such that gg~ l = g~ l g= e. 

Groups occur naturally in many parts of geometry, and we shall men- 
tion a few in subsequent exercises. Basic properties of groups may be 
found, for example, in Birkhoff and MacLane [2]. 

7. Prove that the set 8 of all isometries of E 3 forms a group — with compo- 
sition of functions as the operation. 8 is called the Euclidean group (of 
order 3), or the group of Euclidean motions of E 3 . 



104 EUCLIDEAN GEOMETRY [Chap. Ill 

A subset H of a group G is a subgroup of G provided (1) if g x and g 2 
are in H, then so is g^, (2) if g is in H, so is gr\ and (3) the identity 
element e of G is in #. A subgroup H of G is automatically a group. 

8. Prove that the set 3 of all translations of E 3 and the set 0(3) of all 
orthogonal transformations of E 3 are each subgroups of the Euclidean 
group S. (3) is called the orthogonal group of order 3. Which isometries 
of E 3 are in both these subgroups? 



2 The Derivative Map of an Isometry 

In Chapter I we showed that an arbitrary mapping F: E 3 -*• E 3 has a 
derivative map F* which carries each tangent vector v at p to a tangent 
vector F*(\) at F(p). If F is an isometry, its derivative map is remarkably 
simple. (Since the distinction between tangent vector and point is crucial 
here, we temporarily restore the point of application to the notation.) 

2.1 Theorem Let F be an isometry of E 3 with orthogonal part C. Then 

n(v p ) = (CVW, 

for all tangent vectors \ p to E 3 . 

Verbally: To get F+(v p ), first shift the tangent vector v p to the canoni- 
cally corresponding point v of E 3 , then apply the orthogonal part C of F, 
and finally shift this point C(v) to the canonically corresponding tangent 
vector at F(p) (Fig. 3.2). Thus all tangent vectors at all points p of E 3 are 
"rotated" in exactly the same way by F* — only the new point of application 
F(p) depends on p. 

Proof. Write F = TC as in Theorem 1.7. Let T be translation by a, 
so F(p) = a + C(p). If \ p is a tangent vector to E 3 , then by Definition 
7.4 of Chapter I, F*(v p ) is the initial velocity of the curve t — > F(p + tv). 
But using the linearity of C, we obtain 

F(p + tv) = TC(p + tv) = T(C(p) + tC(v)) = a + C(p) + tC(v) 
= F(p) + tC(v). 




FIG. 3.2 



Sec. 2] THE DERIVATIVE MAP OF AN ISOMETRY 105 

Thus jP*(v p ) is the initial velocity of the curve t — > F(p) + tC(v), which 
is precisely the tangent vector (Cv) F(p ). | 

Expressed in terms of Euclidean coordinates, this result becomes 

F*(E, VjUj) = X) CijVjUi 

i i,i 

where C = (dj) is the orthogonal part of the isometry F, and if Ui is 
evaluated atp, then Ui is evaluated at F(p). 

2.2 Corollary Isometries preserve dot products of tangent vectors. 
That is, if \ p and w p are tangent vectors to E at the same point, and F 
is an isometry, then 

F*(v p )»F*(w p ) = \ p '-w p . 

Proof. Let C be the orthogonal part of F, and recall that C, being an 
orthogonal transformation, preserves dot products in E 3 . By Theorem 2.1, 

F*{y,)*F*(w,) = (Cv) np) -(Cw) F(p) = Cy-Cw 

= VW = Vp'Wp 

where we have twice used Definition 1.3 of Chapter II (dot products of 
tangent vectors). | 

By proving this fundamental corollary and the following theorem, the 
initial result (Theorem 2.1) has largely accomplished its mission. Thus 
it should be safe to drop the point of application from the notation once 
again, and write simply F*(\)»F*(w) = vw. In fancier language, the 
corollary asserts that for each point p, the derivative map F* p at p is an 
orthogonal transformation of tangent spaces (differing from C only by 
the canonical isomorphisms of E 3 ). 

Since dot products are preserved, it follows automatically that derived 
concepts such as norm and orthogonality are preserved. Explicitly, if F 
is an isometry, then || /^*Cv)|| = || v [|, and if v and w are orthogonal, so 
are jP*(v) and F*(w). Thus frames are also preserved: if ci, e 2 , e 3 is a frame 
at some point p of E 3 and F is an isometry, then / r *(e 1 ), F*(e 2 ), F*(e 3 ) 
is a frame at F(p). (A direct proof is easy: e^-ey = 8^, so by Corollary 2.2, 
F*(e<).F*(c y ) = ere,- = 5< y .) 

Assertion (3) of Lemma 1.4 shows how two points uniquely determine a 
translation. We now show that two frames uniquely determine an isometry. 

2.3 Theorem Given any two frames on E 3 , say d, e^, e 3 at the point p 
and fi, f 2 , f 3 at the point q, there exists a unique isometry F of E 3 such that 
F»fe<) = f,forl ^ i£ 3. 

Proof. First we show that there is such an isometry. Let ei, e 2 , e 3 , and 
/i, h, h be the points of E canonically corresponding to the vectors in 



106 EUCLIDEAN GEOMETRY [Chap. Ill 

the two frames. Let C be the unique linear transformation of E 3 such that 
C(e») = fi for 1 ^ i ^ 3. It is easy to check that C is orthogonal. Then 
let T be a translation by the point q - C (p). Now we assert that the isome- 
try F = TC carries the e frame to the f frame. First note that 

F(p) = T(Cp) = q - C(p) + C(p) = q. 
Then using Theorem 2.1 we get 

F*(e t ) = (Ce t ), (p) = (/.-),(,) = (/„■), = f { 
for 1 ^ i ^ 3. 

To prove uniqueness, we observe that by Theorem 2.1 this choice of C 
is the only possibility for the orthogonal part of the required isometry. 
The translation part is then completely determined also, since it must carry 
C(p) to q. Thus the isometry F = TC is uniquely determined. | 

Explicit computation of the isometry in the theorem is not difficult. 
Let e,- = (a a , a i2 , o«) and /< = (6 a , b a , b a ) for 1 ^ i ^ 3. Thus the (or- 
thogonal) matrices A = (a ti ) and £ = (6 l7 ) are the attitude matrices of 
the frames e x , es, e 3 and fi, f 2 , f 3 , respectively. We claim that C in the theo- 
rem (or, strictly speaking, its matrix) is l BA. It suffices to show that 
BA (ei ) = /,-, since this uniquely characterizes C. But using the column- 
vector conventions, we have 

'BA a 12 = '£ U 

\aj W \6 13/ 

that is, 'BA(ei) = /i (the cases i = 2, 3 are similar). Thus C = l BA 
As noted above, 7 1 is then necessarily translation by q — C(p). 



EXERCISES 

1. If T is a translation, then for every tangent vector v show that 7 1 *(v) 
is parallel to v (same Euclidean coordinates). 

2. Prove the general formulas (GF)* = G*F* and (F -1 )* = (f 7 *) -1 
in the special case where F and G are isometries of E 3 . 

3. (a) Let e x , e2, e 3 be a frame at p with attitude matrix A. If F is the 

isometry that carries the natural frame at to this frame, show 
thatF = T p A~ l (A -1 = l A). 
(b) Now let fi, f 2 , f 3 be a frame at q with attitude matrix B. Use Ex- 
ercise 2 to prove the result in the text that the isometry which 
carries the e frame to the f frame has orthogonal part B _1 A. 



Sec. 3] ORIENTATION 107 

4. (a) Prove that an isometry F = TC carries the plane through p or- 

thogonal to q to the plane through F(p) orthogonal to C(q). 
(b) If P is the plane through (£, — 1,0) orthogonal to (0, 1,0) find an 
isometry F = TC such that F(P) is the plane through (1, -2, 1) 
orthogonal to (1 , 0, — 1 ) . 

5. Given the frame ei = (2,2,l)/3, e 2 = (-2,l,2)/3, e 3 = (1, -2,2)/3 

at p = (0, 1,0) and the frame 

f 2 = (l,0,l)/\/2, f 2 = (0,1,0), f 3 = (1,0, -1)/V2 

at q = (3, —1, 1), find the isometry F = TC which carries the 
e frame to the f frame. 



3 Orientation 

We now come to one of the most interesting and elusive ideas in geometry. 
Intuitively, it is orientation that distinguishes between a right-handed glove 
and a left-handed glove in ordinary space. To handle this concept mathe- 
matically, we replace gloves by frames, and separate all the frames on E 3 
into two classes as follows. Recall that associated with each frame ei, e^, e 3 
at a point of E is its attitude matrix A . According to the exercises for Sec- 
tion 1 of Chapter II, 

ei»e 2 X e 3 = det A = ±1. 

When this number is +1, we shall say that the frame e x , e-j, e 3 is positively 
oriented (or, right-handed); when it is —1, the frame is negatively oriented 
(or, left-handed). 
We omit the easy proof of the following facts. 

3.1 Remark (1) At each point of E 3 the frame assigned by the natural 
frame field U u £/ 2 , U 3 is positively oriented. 

(2) A frame ei, e%, e 3 is positively oriented if and only if e x X ^ = e 3 . 
Thus the orientation of a frame can be determined, for practical purposes, 
by the "right-hand rule" given at the end of Section 1 of Chapter Il.Pic- 
torially, the frame (P) in Fig. 3.3 is positively oriented, whereas the frame 
(N) is negatively oriented. In particular, Frenet frames are always positively 
oriented, since by definition B = T X N. 

(3) For a positively oriented frame e x , e 2 , e 3 , the cross products are 

ei = e 2 X e 3 = — e 3 X ^ 

e 2 = e 3 X ei = — e x X e 3 

e 3 = ei X e 2 = — e 2 X ei. 

For a negatively oriented frame, reverse the vectors in each cross product. 



108 



EUCLIDEAN GEOMETRY 



[Chap. Ill 



(P) 




e 2 



FIG. 3.3 



(N) 




ei 



(One need not memorize these formulas — the right-hand rule will give 
them all correctly.) 

Having attached a sign to each frame on E 3 , we next attach a sign to 
each isometry F of E 3 . In Chapter II we proved the well-known fact that 
the determinant of an orthogonal matrix is either +1 or —1. Thus if C is 
the orthogonal part of the isometry F, we define the sign of F to be the 
determinant of C, with notation 

sgn F = det C. 

We know that the derivative map of an isometry carries frames to 
frames. The following result tells what happens to their orientations. 

3.2 Lemma If e x , e2, e 3 is a frame at some point of E 3 and F is an isome- 
try, then 

F*(c0.F*(©2) X /^*(e 3 ) = sgnF ei .e 2 X e 3 . 

Proof. If ej = 23 a JkU k , then by the coordinate form of Theorem 2.1 
we have 

F*(ej) = z2 CikdjkUi 

i,k 

where C = (cij) is the orthogonal part of F. Thus the attitude matrix of 
the frame / p *(e 1 ), F^M, ^*(e 3 ) is the matrix 

(2 cafljk) = (23 c ik Wy) = C l A. 

k k 

But the triple scalar product of a frame is the determinant of its attitude 
matrix, and by definition sgn F = det C. Consequently, 

F*( ei ).F*(e 2 ) X i^M = det (C l A) 

= det (7-det *A = det C-det A 



= sgnjPei«e2 X e 3 . 



I 



This lemma shows that if sgn F = +1, then F* carries positively oriented 
frames to positively oriented frames, and carries negatively oriented to 



Sec. 3] 



ORIENTATION 



109 



negatively oriented frames. On the other hand, if sgn F — — 1, positive 
goes to negative, and negative to positive. 

3.3 Definition An isometry F of E 3 is said to be 

orientation-preserving if sgn F = det C = + 1 
orientation-reversing if sgn F = det C = — 1 
where C is the orthogonal part of F. 

3.4 Example 

(1) Translations. All translations are orientation-preserving. Geometrically 
this is clear, and in fact the orthogonal part of a translation T is just the 
identity mapping i", so sgn T = det / = +1. 

(2) Rotations. Consider the orthogonal transformation C given in Example 
1.2, which rotates E 3 through angle around the z axis. Its matrix is 



COS0 


— sin 





sin 6 


cos 6 











1 



Hence sgn C — det C = +1, so C is orientation-preserving (see Exercise 4). 

(3) Reflections. One can (literally) see reversal of orientation by using a 
mirror. Suppose the yz plane of E 3 is the mirror. If one looks toward that 
plane, the point p = (pi,P2,P3) appears to be located at the point 

R(P) = (~PhP2,Ps) 

(Fig. 3.4). The mapping R so denned is called reflection in the yz plane. 
Evidently it is an orthogonal transformation with matrix 



-1 











1 











1 



Thus R is an orientation-reversing isom- 
etry, as confirmed by the experimental 
fact that the mirror image of a right hand 
is a left hand. 

Both dot and cross product were orig- 
inally defined in terms of Euclidean co- 
ordinates. We have seen that the dot 
product is given by the same formula, 

VW = (52 Vi«i)" (22 «\«t) = 52 v Wi, 




,<°R(p) 



edge view of yz plane 
FIG. 3.4 



ei 


e 2 


e 3 


Vi 


V2 


Vi 


W\ 


102 


w 3 



110 EUCLIDEAN GEOMETRY [Chap. Ill 

no matter what frame ei, e^ e 3 is used to get coordinates for v and w. 
We have almost the same result for cross products, but orientation is now 
involved. 

3.5 Lemma Let d, 62, e 3 be a frame at a point of E 3 . If v = ^ v&i 
and w = ^2 Wiei, then 



v X w = e 



where e = e^^ X e 3 = ±1. 

Proof. It suffices merely to expand the cross product 

v X w = (wid + w 2 e 2 + v 3 e 3 ) X (wiei + W2e<s + w 3 e 3 ) 

using the formulas (3) of Remark 3.1. For example, if the frame is posi- 
tively oriented, for the ei component of v X w, we get 

the* X w 3 e 3 + v 3 e 3 X W2e2 = (v 2 w 3 —v 3 w 2 )ei. 

Since e = 1 in this case, we get the same result from the right side of the 
equation to be proved. | 

It follows immediately that the effect of an isometry on cross products 
also involves orientation. 

3.6 Theorem Let v and w be tangent vectors to E at p. If F is an isOme- 
try of E 3 , then 

F*(v X w) = sgn F F*(v) X F*(w). 
Proof. Write v = ^ i>,-t/,-(p) and w = YL WiUifo). Now let 

e,- = F+(Ui(p)). 
Since F* is linear, 

F*(y) = 2 v i e i an d F*(w) = 2 «*,«,•. 
A straightforward computation using Lemma 3.5 shows that 

F*(v) X F*(w) = e F*(v X w), 
where 

e = ei .e 2 X e 3 = F,(C7i(p)).F* (C7 2 (p)) X F*(t/ 3 (p)). 
But t/i, C7 2 , U 3 is positively oriented, so by Lemma 3.2, e = sgn F. | 



Sec. 3] 



ORIENTATION 



111 



EXERCISES 

1. Prove 

sgn (FG) = sgn F-sgn G = sgn (GF). 
Deduce that sgn F = sgn (F _1 ). 

2. If Ho is an orientation-reversing isometry of E 3 , show that every orien- 
tation-reversing isometry has a unique expression HqF, where F is 
orientation-preserving. 

3. Let v = (3, 1, —1) and w = (-3, -3, 1) be tangent vectors at some 
point. If C is the orthogonal transformation given in Exercise 4 of 
Section 1, check the formula 

C*(v X w) = sgn C C*(v) X C*(w). 

4. A rotation is an orthogonal transformation C such that det C = +1. 
Prove that C does, in fact, rotate E 3 around an axis. Explicitly, given a 
rotation C, show that there exists a number # and points d, d, e 3 with 
e t -»ey = Sij such that (Fig. 3.5) 

C (ei ) = cos # d + sin # e 2 

C(e 2 ) = —sin i? d + cos # d 

C(e.) = e 3 . 

(Hint: The fact that the dimension of E 
is odd means that C has a characteristic 
root + 1, so there is a point p ^ such that 
C(p) =p.) 

5. Let a be a point of E 3 such that || a || = 1. Prove that the formula 

C(p) = a X p +p*a a 

defines an orthogonal transformation. Describe its general effect on E . 

6. Prove 

(a) The set + (3) of all rotations of E 3 is a subgroup of the orthogonal 
group 0(3) (see Ex. 8 of III.l). 

(b) The set 8 + of all orientation-preserving isometries of E is a subgroup 
of the Euclidean group S. 

7. Find a single formula for all isometries of the real line E 1 . Do the same 
for the plane E 2 (use e = ±1). Which of these isometries are orienta- 
tion-preserving? 




FIG. 3.5 



1,2 EUCLIDEAN GEOMETRY [Chap. Ill 

4 Euclidean Geometry 

In the discussion at the beginning of this chapter, we recalled a fundamental 
feature of plane geometry: If there is an isometry carrying one triangle 
onto another, then the two (congruent) triangles have exactly the same 
geometric properties. A close examination of this statement will show 
that it does not admit a proof — it is, in fact, just the definition of "geo- 
metric property of a triangle." More generally, Euclidean geometry can be 
defined as the totality of concepts that are preserved by isometries of 
Euclidean space. For example, Corollary 2.2 shows that the notion of dot 
product on tangent vectors belongs to Euclidean geometry. Similarly 
Theorem 3.6 shows that the cross product is preserved by isometries (ex- 
cept possibly for sign). 

This famous definition of Euclidean geometry is somewhat generous, 
however. In practice, the label "Euclidean geometry" is usually attached 
only to those concepts that are preserved by isometries, but not by arbi- 
trary mappings, or even the more restrictive class of mappings (diffeo- 
morphisms) that possess inverse mappings. An example should make this 
distinction clearer. If a = (ai,a 2 ,a 3 ) is a curve in E 3 , then the various 
derivatives 

a = ( ^ gl ^? d<*z \ „ _ (d 2 ai d 2 a 2 d 2 <x 3 \ 

look pretty much alike. Now, we interpreted Theorem 7.8 of Chapter I 
as saying that velocity is preserved by arbitrary mappings T^: E 3 — >■ E 3 . That 
is, if = F(a), then /? = F*(a ). But it is easy to see that acceleration 
is not preserved by arbitrary mappings. For example, if a(t) = (2,0,0) and 
F = (x\y,z), then a" = 0; hence F+{a") = 0. But p = F(a) has the 
formula 0(0 = (* 2 ,0,0), so 0" = 2C/i. Thus in this case, = F(a), but 
$" 5^ F*(a"). We shall see in a moment, however, that acceleration is 
preserved by isometries. 

For this reason, the notion of velocity belongs to the calculus of Euclid- 
ean space, while the notion of acceleration belongs to Euclidean geometry. 
In this section we examine some of the concepts introduced in Chapter II 
and prove that they are, in fact, preserved by isometries. (We leave largely 
to the reader the easier task of showing that they are not preserved by 
diffeomorphisms. ) 

Recall the notion of vector field on a curve (Definition 2.2 of Chapter 
II). Now if Y is a vector field on a: I -> E 3 , and F: E 3 — ► E 3 is any mapping, 
then Y = F*(Y) is a vector field on the image curve a = F(a). In fact, 
for each /in/, Y(t) is a tangent vector to E 3 at the point a (t). But then 
Y(t) = F*(Y(t)) is a tangent vector to E 3 at the point F(a(t)) = a(t). 



Sec. 4] 



EUCLIDEAN GEOMETRY 



113 




FIG. 3.6 



(These relationships are illustrated in Fig. 3.6.) Isometries preserve the de- 
rivatives of such vector fields. 

4.1 Corollary Let Y be a vector field on a curve a in E 3 , and let F be 
an isometry of E 3 . Then Y = F*(F) is a vector field on a = F(a), and 

f = F+iY*). 

Proof. We compute F*(Y') and Y starting from the expression 

Y = Z ViUt 

for Y in terms of its Euclidean coordinate functions. To differentiate such 
a vector field Y, one simply differentiates its Euclidean coordinate functions, 
so 

F' = Zf V, 
Thus by the coordinate version of Theorem 2.1, we get 



On the other hand, 



F*(Y') =T,^U i . 



Y = F*(Y) = T.CiiViUi. 



But each c t _, is constant, being by definition an entry in the matrix of the 
orthogonal part of the isometry F. Hence 



Thus the vector fields F*(Y') and Y' are the same. 



I 



We claimed earlier that isometries preserve acceleration: If a = F(a), 
where F is an isometry, then a" = F*(a"). This is an immediate conse- 



"4 EUCLIDEAN GEOMETRY [Chap. Ill 

quence of the preceding result, for if we set Y = a, then by Theorem 7.8 
of Chapter I, Y = a; hence 

5 " =Y' = F*(Y') = F*(aT). 

Now we show that the Frenet apparatus of a curve is preserved by 
isometries. This is certainly to be expected on intuitive grounds, since a 
rigid motion ought to carry one curve into another that turns and twists 
in exactly the same way. And this is what happens when the isometry is 
orientation-preserving . 

4.2 Theorem Let be a unit-speed curve in E 3 with positive curvature, 
and let = F(0) be the image curve of under an isometry F of E 3 . Then 

k = k f = F*(T) 

f = sgn Ft N = F*(N) 

B = sgn FF*(B) 

where sgn F = ±1 is the sign of the isometry F. 

Proof. Note that is also a unit-speed curve, since 

ii in = ii^goii = \\ft\\ = i. 

Thus the definitions in Section 3 of Chapter II apply to both and /3, so 

f = ff = F*(/3') = F m (T). 

Since F* preserves both acceleration and norms, it follows from the 
definition of curvature that 

* = H0-1 = imonn = ii0" ii = *■ 

To get the full Frenet frame, we now use the hypothesis k > (which 
implies Z > 0, since R = k). By definition, N = fi"/n; hence using preceding 
facts, we find 

K K \ K / 

It remains only to prove the interesting cases B and t. Since the defini- 
tion B = T X N involves a cross product, we use Theorem 3.6 to get 

B = f X N = F*(T) X F*(N) = sgn FF*(T X N) = 8gnFF*(B). 

The definition of torsion is essentially r = —B'»N = B-N'. Thus using 
the results above for B and N, we get 

f = B.N' = sgn F F*(B)*F*(N') = sgn F B-N' = sgn Ft. | 

The presence of sgn F in the formula for the torsion of F{fi) shows that 



Sec. 4] 



EUCLIDEAN GEOMETRY 



115 



the torsion of a curve gives a more subtle description 
of the curve than has been apparent so far. The sign 
of t measures the orientation of the twisting of the 
curve. If F is orientation-reversing, the formula f = 
— t proves that the twisting of the image of curve 
F(fi) is exactly opposite to that of itself. 
A simple example will illustrate this reversal. 

4.3 Example Let /3 be the unit-speed helix 

|8(s) = (cos -, sin - , -), 
\ c c c/ 




FIG. 3.7 



gotten from Example 3.3 of Chapter II by setting 

a = b = 1 ; hence c = \/2- We know from the general formulas for helices 

that k = t = \. Now let R be reflection in the xy plane, so R is the isom- 

etry R{x,y,z) = (x,y,—z). Thus the image curve (8 = R(p) is the mirror 

image 



/8(«) = 



(s . s s\ 
cos - , sin - , — - J 
c c c/ 



of the original curve. One can see in Fig. 3.7 that the mirror has its usual 
effect: and fi twist in opposite ways — if /3 is "right-handed," then /3 is 
"left-handed." (The fact that is going up and j§ down is, in itself, irrele- 
vant.) Formally: The reflection R is orientation-reversing; hence the 
theorem predicts k = k = \ and f = — t = —\. Since /8 is just the helix 
gotten in Example 3.3 of Chapter II by taking a = 1 and b = — 1, this 
may be checked by the general formulas there. 



EXERCISES 

1. Let F = TC be an isometry of E 3 , /3 a unit speed curve in E 3 . Prove 

(a) If /3 is a cylindrical helix, then F(f3) is a cylindrical helix. 

(b) If /3 has spherical image ft, then F(fi) has spherical image C(j8). 

2. Let Y = (t, 1 — t 2 , 1 + t 2 ) be a vector field on the helix 

a(t) = (cos t, sin t, 2t), 
and let C be the orthogonal transformation 

-1 

C = | 1A/2 -1/V2 
1/V2 1/V2/ 



"* EUCLIDEAN GEOMETRY [Chap. Ill 

Compute a = C(a) and Y = C*(F), and check that 

C*(F / ) = ?', C*(a") = a", F'-a" = ?'.«". 

3. Sketch the triangles in E 2 that have vertices 

A i: (3, 1), (7, 1), (7, 4) A 2 : (2, 0), (2, 5), (-f, ^) 

Show that these triangles are congruent by exhibiting an isometry F 
that carries Ai to A 2 . (Hint: the orthogonal part of F is not altered if 
the triangles are translated.) 

4. If F: E —> E 3 is a mapping such that F* preserves dot products, show 
that F is an isometry. (Hint: Use Ex. 11 of II.2.) 

5. Let F be an isometry of E 3 . For each vector field V let V be the vector 
field such that F*( F(p) = 7(F(p) ) for all p. P rove that isometries pre- 
serve covariant derivatives; that is, show y v W = VrW. 



5 Congruence of Curves 

In the case of curves in E 3 , the general notion of congruence takes the 
following form. 

5.1 Definition Two curves a ,0: I — > E are congruent provided there 
exists an isometry F of E 3 such that ft = F(a); that is, /?(£) = F(a(t)) 
for all tin I. 

Intuitively speaking, congruent curves are the same except for position 
in space. They represent trips at the same speed along routes of the same 
shape. For example, the helix a(t) = (cos t, sin t, t) spirals around the 
z axis in exactly the same way the helix (i(t) = (t, cos t, sin t) spirals 
around the x axis. Evidently these two curves are congruent, since if F 
is the isometry such that 

F(Vu Pi, P») = (PfcPi, P*), 

then F (a) = p. 

To decide whether given curves a and /8 are congruent, it is hardly 
practical to try all the isometries of E 3 to see if there is one that carries 
a to |8. What we want is a description of the shape of a unit-speed curve so 
accurate that if a and /? have the same description, then they must be 
congruent. The proper description, as the reader will doubtless suspect, 
is given by curvature and torsion. To prove this we need one preliminary 
result. 

Curves whose congruence is established by a translation are said to be 
parallel. Thus curves a, /3: / — * E are parallel if and only if there is a point 



Sec. 5] 



CONGRUENCE OF CURVES 



117 



p in E 3 such that /3(s) = a(s) + p for all s in /, or, in functional nota- 
tion, /3 = a -+- p. 

5.2 Lemma Two curves a,/3: / — »• E 3 are parallel if their velocity vectors 
a (s) and (s) are parallel for each s in /. In this case, if a(so) = P(s ) 
for some one s in /, then a = 0. 

Proof. By definition, if a (s) and /8 (s) are parallel, they have the same 
Euclidean coordinates. Thus 



— f ") = — M 



for 1 < i < 3 



where a» and /3i are the Euclidean coordinate functions of a and /S. But 
by elementary calculus, the equation doti/ds = dPi/ds implies that there is a 
constant pi such that ft = a t + Pi- Hence /3 = a + p. Furthermore, if 
a(so) = (8(so), we deduce thatp = 0; hence a = 0. | 

5.3 Theorem If a, /3: / — » J57 are unit-speed curves such that *c« = k# 
and r a = ±7-0, then a and /3 are congruent. 

Proof. There are two main steps: 

(1) Replace a by a suitably chosen congruent curve F(a). 

(2) Show that F(a) = |9 (Fig. 3.8). 

Our guide for the choice in (1 ) is Theorem 4.2. Fix a number, say 0, 
in the interval /. If r a = t/j, then let F be the (orientation-preserving) 
isometry that carries the Frenet frame T a (0), N a (0), B a (0) of a at a (0) 
to the Frenet frame T(0), N(0), B(0), of ■ at /3(0). (The existence of this 
isometry is guaranteed by Theorem 2.3.) Denote the Frenet apparatus of 
a = F(a) by k, f, f, N, B; then it follows immediately from Theorem 4.2 
and the information above that 



«(0) = 0(0) 

K = Kp 
f = 70 



«m 




T(0) 


= T(0) 


N(0) 


= N(0) 


8(0) 


= B(0). 




a = F{aV) 
4 



(t) 



5(0) = /3(0> 




FIG. 3.8 



118 EUCLIDEAN GEOMETRY [Chap. Ill 

On the other hand, if r a = —t$, we choose F to be the (orientation- 
reversing) isometry that carries T a (0), N a (0), B a (0) at a(0) to the frame 
T(0), N(0), —B(0) at /3(0). (Frenet frames are positively oriented; 
hence this last frame is negatively oriented : This is why F is orientation- 
reversing.) Then it follows from Theorem 4.2 that the equations (f) hold 
also for a = F(a) and £. For example, 

B(0) = -F*(B a (0)) = 5(0). 

For step (2) of the proof, we shall show T = T; that is, the unit tangents 
of a = F(a) and are parallel at each point. Since a(0) = /3 (0 ) , it will 
follow from Lemma 5.2 that F(a) = /?. On the interval /, consider the 
real- valued function/ = T»T + N»N + B'B. Since these are unit vector 
fields, the Schwarz inequality (Section 1, Chapter II) shows that 

T-T g: 1; 

furthermore f »T = 1 if and only if f = T. Similar remarks hold for the 
other two terms in /. Thus it suffices to show that f has constant value 3. 
By (t),/(0) = 3. Now consider 

/ = T'-T + T-T' + N'-N + N-N' + B'-B + B-B'. 

A simple computation completes the proof. Substitute the Frenet formu- 
las in this expression and use the equations H = k, f = t from (J). The 
resulting eight terms cancel in pairs, so / =0, and / has, indeed, constant 
value 3. | 

Thus a unit-speed curve is determined but for position in E by its curvature 
and torsion. 

Actually the proof of Theorem 5.3 does more than establish that a and /3 
are congruent; it shows how to compute explicitly an isometry carrying 
a to )8. We illustrate this in a special case. 

5.4 Example Consider the unit-speed curves a, /3: R — » E such that 



( \ ( s . s s\ 

a\s) = I cos - , sin- , - ] 

\ c c c/ 

/3(s) = I cos - , sin - , ) 

\ c c c/ 



where c = \/2. Obviously these curves are congruent by means of a 
reflection — they are the helices considered in Example 4.3 — but we shall 
ignore this in order to describe a general method for computing the re- 
quired isometry. According to Example 3.3 of Chapter II, a and /8 have the 
same curvature, n a = \ = Kp; but torsions of opposite sign, r a = \ = — 7> 
Thus the theorem predicts congruence by means of an orientation-revers- 



Sec. 5] CONGRUENCE OF CURVES 119 

ing isometry F. From its proof we see that F must carry the Frenet frame 

T a (0) = (0,a, a) 

N a (0) = (-1,0,0) 

B a (0) = (0, -a, a) 
to the frame 

T,(0) = (0, a, -a) 

N (O) = (-1,0,0) 

-B (O) = (0, -a, -a) 

where a = l/-\/2. (These explicit formulas also come from Example 3.3 of 
Chapter II.) By the remark following Theorem 2.3, the isometry F has 
orthogonal part C = l BA, where A and B are the attitude matrices of the 
two frames above. Thus 

a a\ /l 0\ 

-1 0=01 

-a a/ \0 -l y 

since a = l/\/2- These two frames have the same point of application 
a(0) = |8(0) = (1, 0, 0). But C does not move this point, so the transla- 
tion part of F is just the identity map. Thus we have (correctly) found 
that the reflection F = C carries a to 0. 

From the viewpoint of Euclidean geometry, two curves in E 3 are "the 
same" if they differ only by an isometry of E 3 . What, for example, is a helix? 
Not just a curve that spirals around the z axis as in Example 3.3 of Chapter 
II, but any curve congruent to one of these special helixes. One can give 
general formulas, but the best characterization follows. 

5.5 Corollary Let a be a unit speed curve in E 3 . Then a is a helix if and 
only if both its curvature and torsion are nonzero constants. 

Proof. For any numbers a > and 6^0, let &,& be the special helix 
given in Example 3.3 of Chapter II. If a is congruent to |S„, 6 , then (changing 
the sign of b if necessary) we can assume the isometry is orientation- 
preserving. Thus, a has curvature and torsion 




a 2 + b 2 a 2 + 6 2 " 

Conversely, suppose a has constant nonzero k and t. Solving the pre- 
ceding equations, we get 



120 EUCLIDEAN GEOMETRY [Chap. Ill 



* 1, 

a = -7— — -„ b = 



K 2 + T 2 K 2 + T 2 

Thus a and &,& have the same curvature and torsion, hence they are 
congruent. | 

Our results so far demand unit speed, but it is easy to weaken this re- 
striction. 

5.6 Corollary Let a, /3: / — > E be arbitrary-speed curves. If 

Va = Vp > 0, K a = K0 > 0, and T a — ±T0, 

then the curves a and /3 are congruent. 

Proof. Let 5 and $ be unit-speed reparametrizations of a and (3, both 
based at, say, t = 0. Since a and jS have the same speed function, it follows 
immediately that they also have the same arc length function s = s(t) 
and hence the same inverse function t = t(s). But since 

Ka = K(S and T a = ± T0, 

we deduce from the general definitions of curvature and torsion in that Sec- 
tion 4 of Chapter II 

Ka(s) = K a (t(s)) = Kfi(t(s)) = K$(s) 

Ta(s) = T a (t(s)) = ±Tfi(t(s)) = ±T^(s). 

Hence the congruence theorem (5.3) shows that a and are congruent 
— say, F(a) = j8. But then the same isometry carries a to /?, since 

F(a(t)) = F(a(s(t))) = F(p(s(t))) = F(p(t)). | 

The theory of curves we have presented applies only to regular curves 
with positive curvature, because only for such curves is it generally pos- 
sible to define the Frenet frame field. However, a completely arbitrary 
curve a in E can be studied by means of an arbitrary frame field on a, 
that is, any three unit-vector fields Ei, E 2 , E 3 on a that are orthogonal at 
each point (Fig. 3.9). 




FIG. 3.9 



Sec. 5] CONGRUENCE OF CURVES 121 

For example, the congruence theorem 5.3 can easily be extended to arbi- 
trary curves. 

5.7 Theorem Let a, /?: / — > E 3 be arbitrary curves, and let E h E 2 , E 3 be 
a frame field on a; F u F 2 , F 3 a frame field on 0. If 

(1) a''Ei = p'-Fi (l£i ^ 3) 

(2) Ei-Ei = F/.Fj (1 ^ i,j S 3), 
then a and are congruent. 

Proof. We need only generalize the argument in Theorem 5.3. Fix a 
number, say 0, in /. Then let F be the isometry that carries 

E 1 (0), E 2 (0), E 3 (0) to F t (0), F 2 (0), F 3 (0). 

Since F* preserves dot products, it follows that Ei = F*(Ei) (1 ^ i ^ 3) 
is a frame field on a = F(a). Since F* preserves velocities and derivatives 
of vector fields as well, we deduce 

a(0) = 0(0) a' -Ei = fl'-Fi 

Ei(0) = F,-(0) E/.Ej = Ft'-Fj for 1 ^ i,j ^ 3. 

This last equation means we can write £/ = 23 QijEj and F/ = 23 G^i 
with the same coefficient functions a,-,-. Note that a ti -f a>i = 0. (Differ- 
entiate Ei*Ej = 8ij.) If / = 23 Ei'Fi, we then prove/ = 3 just as before, 
since 

/' = 23 (^/-*\- + Ei'Fi') = 23 («<y + aii)E r Fi = 0. 

Thus Ei = Fi (parallelism!) and it follows from ({) that 

«' = E a'.-E,- & and /?' = 23 p'.Fi F { 

are parallel at each point. But a(0) = /3(0); hence by Lemma 5.2, 

F(a) = a = |8. | 

We shall need this degree of generalization in Section 8 of Chapter VI. 
A more elegant (but slightly less general) version of this theorem is given 
in Exercise 3. 



EXERCISES 

1. Given a curve a = (a 1} a 2 , a 3 ): I — > E 3 , prove that 0: / — » E 3 is con- 
gruent to a if and only if /3 can be written 

0(0 = P + ati(t)ei + a 2 (0e 2 + a 3 (t)e 3 , 
where e»»Cj = 5i_,. 



122 EUCLIDEAN GEOMETRY [Chap. Ill 

2. Let a be the curve in Example 4.4 of Chapter II. Find a (congruent) 
curve of the form 7(0 = (at, bt , ct ) and an isometry F such that 
F(a) = y. 

3. Let Ei, E 2 , E3 be a frame field on E with dual forms 0; and connection 
forms a>ij. Prove that two curves a, /3: / — » E are congruent if we have 
Bi(a) = Oi(fi') and «,-,■(«') = coaifi') for 1 ^ ij ^ 3. 

4. Show that the curve 

p(t) = (t + V3 sin t, 2 cos t, y/Z t - sin t) 

is a helix by computing its curvature and torsion. Find a helix a of 
the form (a cos t, a sin £, 60 and an isometry F such that F(a) = /3. 

5. Let a,(i: I — *■ E be congruent curves with k > 0. Show that there is 
only one isometry F such that F(a) = (8 — unless r = 0, in which case 
there are exactly £wo. 

6. (Continuation). Find the two isometries carrying the parabola a(t) = 
(v%, £ 2 , 0) to the parabola 0(t) = (-t,t,t 2 ). 

7. If j8 is a unit-speed curve in E , then every unit-speed reparametriza- 
tion j8 of j8 has the form j8(s) = j8(±s + s ). If (3 and |8 are congruent, 
this represents a symmetry in the common route of /3 and j§. Prove that 
helical routes are completely symmetric. Explicitly, show that the 
helix |8 in Example 3.3 of Chapter II is congruent to every unit-speed 
reparametrization /3 by explicitly finding the isometry F = TC such 
thatF(0) = /3. 

8. Two curves a : I — * E and £ : / — » E have congruent routes provided 
there is an isometry F such that F(a) is a reparametrization of 0. 

(a) Show that unit-speed curves a and j8 have congruent routes if 
and only if there is a number s such that K a (s) = Kp(es + so) and 
t„(s) = dzTp(es + So), where e is either +1 or —1. 

(b) If a is the curve in Ex. 2 of II.4 show a and = (V, e~'/2, 
have congruent routes. Exhibit the isometry F = TC and the re- 
parametrization needed to fulfill the definition. 

The following three exercises deal with curves in E 2 . 

9. Given any differentiate f function k on an interval /, prove that there 
is a unit-speed curve a in E 2 such that k is the curvature function of a. 
(Hint: Find an integral formula for a by reversing the order of results 
in Ex. 8 of II.3.) 

10. Find plane curves — in any convenient parametrization — for which 

t Even if k is merely continuous, we obtain a twice-differentiable curve. Sim- 
ilar results can be proved for curves in E 3 using systems of ordinary differential 
equations. See Willmore [3]. 



Sec. 6] SUMMARY 123 

(a) k(s) = 1/(1 + s 2 ), (b) k(s) = 1/s (s > 0), where s is the 
arc length. 

11. Prove that two unit-speed curves a and in E 2 are congruent if and 
only if K a = ±K£. 



6 Summary 

The basic result of this chapter is that an arbitrary isometry of Euclidean 
space can be uniquely expressed as an orthogonal transformation followed 
by a translation. Its main consequences are that the derivative map of an 
isometry F is at every point essentially just the orthogonal part of F, and 
that there is a unique isometry which carries any one given frame to 
another. Then it is a routine matter to test the concepts introduced earlier 
and discover which belong to Euclidean geometry, that is, which are pre- 
served by isometries of Euclidean space. Finally, we proved an analogue 
for curves of the well-known "side-angle-side," "side-side-side" theorems 
on triangles from elementary plane geometry. Namely, we showed that 
curvature and torsion (and speed) provide a necessary and sufficient con- 
dition for two given curves to be congruent. Furthermore, the required 
isometry can be explicitly computed. 



CHAPTER 



IV 



Calculus on a Surface 



This chapter begins with the definition of a surface in E 3 and with some 
standard ways to construct surfaces. Although this concept is a more-or- 
less familiar one, it is not as widely known as it should be that each surface 
has a differential and integral calculus strictly comparable with the usual 
calculus on the Euclidean plane E 2 . The elements of this calculus — func- 
tions, vector fields, differential forms, mappings — belong strictly to the 
surface and not to the Euclidean space E 3 in which the surface is located. 
Indeed, we shall see in the final section that this calculus survives un- 
damaged when E is removed, leaving just the surface and nothing more. 



1 Surfaces in E 3 

A surface in E is, to begin with, a subset of E , that is, a certain collection 
of points of E 3 . Of course, not all subsets are surfaces: We must certainly 
require that a surface be smooth and two-dimensional. We shall express 
this requirement in mathematical terms by the next two definitions. 

1.1 Definition A coordinate patch x: D — > E is a one-to-one regular 
mapping of an open set D of E 2 into E . 

The image x (D ) of a coordinate patch x — that is, the set of all values of 
x — is a smooth two-dimensional subset of E (Fig. 4.1). Regularity (Defi- 
nition 7.9 of Chapter I), for a patch as for a curve, is a basic smoothness 
condition; the one-to-one requirement is included to prevent x(D) from 
cutting across itself. Furthermore, in order to avoid certain technical 
difficulties (Example 1.7), we shall sometimes use proper patches, those 
for which the inverse function x -1 : x(D) — ■> D is continuous (that is, has 

124 



Sec. 1] 



SURFACES IN E 3 



125 





D 






















FIG. 4.1 

continuous coordinate functions). If we think of D as a thin sheet of rubber, 
we can get x (D ) by bending and stretching D in a not too violent fashion. 
To construct a suitable definition of surface we start from the rough 
idea that any small enough region in a surface M resembles a region in the 
plane E 2 . The discussion above shows that this can be stated somewhat 
more precisely as: Near each of its points, M can be expressed as the image of 
a proper patch. (When the image of a patch x is contained in M, we say 
that x is a patch in M. ) To get the final form of the definition, it remains 
only to define a neighborhood 91 of p in M to consist of all points of M whose 
Euclidean distance from p is less than some number e > 0. 

1.2 Definition A surface in E 3 is a subset M of E 3 such that for each point 
p of M there exists a proper patch in M whose image contains a neighbor- 
hood of p in M (Fig. 4.2). 

The familiar surfaces used in elementary calculus satisfy this definition; 
for example, let us verify that the unit sphere 2 in E 3 is a surface. By defi- 
nition, 2 consists of all points at unit distance from the origin — that is, all 
points p such that 

Up 11= (pi 2 + v? + P 3 2 ) 1/2 = i. 

To check the definition above, we start by finding a proper patch in 2 
covering a neighborhood of the north pole (0, 0, 1). Note that by dropping 




FIG. 4.2 



126 



CALCULUS ON A SURFACE 



[Chap. IV 




(u, v) *-* (u, v, 0) 



FIG. 4.3 



each point (g x , q 2 , qs) of \the northern hemisphere of 2 onto the xy plane 
at (qi, q 2 , 0) we get a one-to-one correspondence of this hemisphere with 
a disc D of radius 1 in the xy plane (see Fig. 4.3 ) . If we identify this plane 
with E 2 by means of the natural association (q lf q 2 , 0) <-> (qi, q 2 ), then D 
becomes the disc in E 2 consisting of all points (u, v) such that u 2 + v < 1. 

Expressing this correspondence as a function on D we find the formula 

x(u, v) = (u, v, y/l — v? — v 2 ). 

Thus x is a one-to-one function from D onto the northern hemisphere 
of 2. We claim that x is a proper patch. The coordinate functions of x are 
differentiable on D, so x is a mapping. To show that x is regular, we com- 
pute its Jacobian matrix (or transpose) 



fdu <to dp 

du du du 

du dv df 

\dv dv dv 



1 



1 



where / = \/l — u 2 — v 2 . Evidently the rows of this matrix are always 
linearly independent, so its rank at each point is 2. Thus by the criterion 
following Definition 7.9 of Chapter I, x is regular, and hence is a patch. 
Furthermore x is proper, since its inverse function x _1 : x (D ) — > D is given 
by the formula 

x _1 (pi, P2, pa) = (pi, pa), 

hence is certainly continuous. Finally we observe that the patch x covers 
a neighborhood of p = (0, 0, 1 ) in S. Indeed it covers a neighborhood of 
every point q in the northern hemisphere (Fig. 4.4). 

In a strictly analogous way, we can find a proper patch covering each 



Sec. 1] SURFACES IN E 3 127 




FIG. 4.4 

of the other five coordinate hemispheres of 2, and thus verify, by Definition 
1.2, that 2 is a surface. Our real purpose here has been to illustrate Defi- 
nition 1.2 — we shall soon find a much quicker way to prove (in particular) 
that spheres are surfaces. 

The argument above shows that if / is any differentiate real-valued 
function on an open set D in E 2 , then the function x: D — > E such that 

x(u, v) = (w, v,f(u, v)) 

is a proper patch. We shall call patches of this type Monge patches. 

We turn now to some standard methods of constructing surfaces. Note 
that the image M — x(D) of just one proper patch automatically satisfies 
1.2; M is then called a simple surface. (Thus Definition 1.2 says that any 
surface in E can be constructed by gluing together simple surfaces.) 

1.3 Example The surface M : z = f(x, y). Every differentiable real- 
valued function / on E 2 determines a surface M in E 3 : the graph of /, 
that is, the set of all points of E 3 whose coordinates satisfy the equation 
z = f(x, y). Evidently M is the image of the Monge patch 

x(u,v) = (u, v,f(u, v)); 

hence by the remarks above, M is a simple surface. 

If g is a real-valued function on E 3 and c is a number, denote by M: 
g = c the set of all points p such that g(p) = c. For example, if g is a tem- 
perature distribution in space, then M: g = c consists of all points of 
temperature c. There is a simple condition that tells when such a subset of 
E is a surface. 

1.4 Theorem Let g be a differentiable real-valued function on E 3 , and c 
a number. The subset M: g(x, y, z) = c of E 3 is a surface if the 
differential dg is not zero at any point of M. 

(In Definition 1.2 and in this theorem we are tacitly assuming that M 
has some points in it; thus the equation x 2 + y + z 2 = —1, for example, 
does not define a surface. ) 



128 



CALCULUS ON A SURFACE 



[Chap. IV 




K3> 

— ^(Pi , Pi , 0) «-» (pi , 1h) 



FIG. 4.5 



Proof. All we do is give geometric content to a famous result of advanced 
calculus — the implicit function theorem. If p is a point of M , we must find 
a proper patch covering a neighborhood of p in M (Fig. 4.5 ) . Now 

dx dy dz 

Thus the hypothesis on dg is equivalent to assuming that at least one of 
these partial derivatives is not zero at p, say (dg/dz) (p) ^ 0. In this case, 
the implicit function theorem says that nearp the equation g(x, y, z) = c 
can be solved for z. More precisely, it asserts that there is a differentiable 
real-valued function h defined on a neighborhood D of (pi, pi) such that 

(1) For each point (u, v) in D, the point (u, v, h(u, v)) lies in M; that 
is, g(u, v, h(u, v)) — c. 

(2) Points of the form (u, v, h(u, v)), with (u, v) in D, fill a neighbor- 
hood of p in M . 

It follows immediately that the Monge patch x: D — * E 3 such that 

x(w, v) = (u, v,h(u,v)) 

satisfies the requirements in Definition 1.2. Since p was an arbitrary point 
of M , we conclude that M is a surface. | 

When M: g = c is a surface, M is said to be defined implicitly by the 
equation g — c. It is now very easy to prove that spheres are surfaces. The 
sphere S in E 3 of radius r > and center c = (c 1} c 2 , C3) is the set of all 
points at distance r from c. If g = 2 (a:* — c») 2 , then S is defined impli- 



Sec. 1] 



SURFACES IN E 3 



129 



citly by the equation g = r . Now dg = 
22 (a;* — Ci)d,Xi, hence dg is zero only at 
the point c, which is not in 2. Thus 2 is 
a surface. 

Using this theorem and the notion of 
curve defined on page 20 we derive two 
well-known types of surfaces. 




FIG. 4.6 



1.5 Example Cylinders. As a line L, 
perpendicular to a plane P, moves along 

a curve C in P, it sweeps out a cylinder. For definiteness, let P be, say, 
the xy plane, so that L is always parallel to the z axis as in Fig. 4.6. If the 
curve C is given by 

C:f(x,y) = c inE 2 , 

let J be the function on E such that J(pi, p 2 , P3) = f(pi, P2). Then the 
resulting cylinder is evidently given by 

M:J(x,y,z) = c inE 3 . 

The definition of curve on page 20 requires that at each point of C either 
df/dx or df/dy is nonzero. Since 

|(pi,P2,Pa) = %(puP»), 

and similarly for d/dy, it follows that dg is never zero at a point of M . 
Thus M is a surface. 

When C is a circle, we obtain a circular cylinder M : x 2 + y — r 2 in E 3 . 
In Example 1.5 we constructed a surface essentially by translating a 
curve; now we get one by rotating a curve. 

1.6 Example Surfaces of revolution. Let C be a curve in a plane P, 
and let A be a line in P which does not meet C. If this profile curve C is 
revolved around the axis A, it sweeps out a surface of revolution M in E 3 . 
We now check, using Theorem 1.4, that M really is a surface. For sim- 
plicity, assume that P is a coordinate plane and A is a coordinate axis — 
say the xy plane and x axis, respectively. Since C must not meet A, we 
assume it is in the upper half (y > 0) of the xy plane. As C is revolved, 
each point (#1, q 2 , 0) of C gives rise to a whole circle of points 

(qi, #2 cos v, q 2 sin v) in M, for ^ v ^ 2ir. 

Put in reverse, a point p = (pi, p 2 , ps) is in M if and only if the point 



130 



CALCULUS ON A SURFACE 



[Chap. IV 



P = (Pi, VP2 2 + P3 2 , 0) 

is in C (Fig. 4.7). 

If the profile cur ve is C: f(x, y) = c, we define a function g on E 3 by 
d(x, V, z) = f(x, VV + z 2 )- Then the argument above shows that the re- 
sulting surface of revolution is exactly M: g(x, y, z) = c. Using the chain 
rule, it is not hard to show that dg is never zero on M , so M is a surface. 

The circles in M generated, under revolution, by each point of C are 
called the parallels of M, the different positions of C as it is rotated are 
called the meridians of M. This terminology derives from the geography of 
the sphere; however, a sphere is not a surface of revolution as defined above. 
Its profile curve must twice meet the axis of revolution, so two "parallels" 
reduce to single points. To simplify the statements of subsequent theorems 
we use a slightly different terminology in this case; see Exercise 12. 

The necessity of the properness condition on the patches in Definition 
1.2 is shown by the following example. 

1.7 Example Suppose that a rectangular strip of tin is bent into a fig- 
ure-8, as in Fig. 4.8. The configuration M which results does not satisfy 
our intuitive picture of what a surface should be, for along the axis A, M 
is not like the plane E 2 but is instead like two intersecting planes. To ex- 
press this construction in mathematical terms, let D be the rectangle 
— ir < u < tt, < ^<linE and define x: D — > E 3 by x(u, v) = (sin u, 
sin 2u, v). It is easy to check that x is a patch, but its image M = \(D) is 




FIG. 4.7 



E* 



D 





FIG. 4.8 



Sec. 1] 



SURFACES IN E 3 



131 



?£> *e>" 




FIG. 4.9 




FIG. 4.10 

not a surface: x is not a proper patch. Continuity fails for x _1 : M — > D 
since, roughly speaking, to restore M to D, x _1 must tear M along the axis 
A (the z axis of E ). 

By Example 1.6, the familiar torus of revolution T is a surface (Fig. 4.19). 
With somewhat more work, one could construct double toruses of various 
shapes, as in Fig. 4.9. By adding "handles" and "tubes" to existing surfaces 
one can — in principle, at least — construct surfaces of any desired degree 
of complexity (Fig. 4.10). 



EXERCISES 

1. None of the following subsets M of E 3 are surfaces. At which points p 
is it impossible to find a proper patch in M that will cover a neighbor- 
hood of p in M? (Sketch M — formal proofs not required.) 

(a) Cone M: z 2 = x 2 + y 2 . 

(b) Closed disc M: x 2 + y 2 ^ 1, z = 0. 

(c) Folded plane M : xy = 0, x ^ 0, y ^ 0. 

2. A plane in E 3 is a surface M : ax + by + cz = d, where the numbers a, 
b, c are necessarily not all zero. Prove that every plane in E 3 may be 
described by a vector equation as on page 60. 

3. Sketch the general shape of the surface M : z = ax 9 + by 2 in each of 
the cases: 



132 CALCULUS ON A SURFACE [Chap. IV 

(a) a > b > (c) a > b = 

(b) a > > b (d) a = b = 0. 

4. In which of the following cases is the mapping x: E 2 — > E 3 a patch? 

(a) x(u,v) = (v, uv, v). (c) x(w, v) = (u, u 2 , v + y 3 ). 

(b) x(w, y) = (w 2 , w 3 , y). (d) x(u, v) = (cos 2iru, sin 2ru, v). 
(Rec,all that x is one-to-one if and only if x(u, v) = x(u h Vi) implies 
(u, v) = (U U Vi).) 

5. (a) Prove that M: (x 2 + y 2 f + 3z 2 = 1 is a surface. 

(b) For which values of c is M : z(z — 2) + xy = c a surface? 

6. Determine the intersection z = of the monkey saddle 

M: z = f(x, y), f = y 3 - Zyx 2 , 

with the xy plane. On which regions of the plane is/> 0?/ < 0? 
How does this surface get its name? 

7. Let x: D— >E be a mapping, with 

x(u, v) = (x 1 (u,v),x 2 (u,v),xz(u,v)). 

(a) Prove that a point p = (pi, P2, Ps) of E 3 is in the image x(D) if 
and only if the equations 

Pi = Xi(u, v) pi = x 2 (u, v) p 3 = x 3 (u, v) 

can be solved for u and v, with (u, v) in D. 

(b) If for every point p in x(D) these equations have the unique 
solution: u = /i(pi, p 2 , Pz), v = / 2 (pi, Pi, Ps), with (w, v) in D, 
prove that x is one-to-one and that x -1 : x (D ) -^ D is given by the 
formula 

x _1 ( P ) = (A(p),/.(p)). 

8. Let x: D — > E 3 be the function given by 

x(w, v) = (w , W, V ) 

on the first quadrant D : u > 0, v > 0. Show that x is one-to-one and 
find a formula for its inverse function x _1 : x(D) — > D. Then prove 
that x is a proper patch. 

9. Let x: E 2 — > E 3 be the mapping 

x(u, v) = {u -{- v, u — v, uv). 

Show that x is a proper patch, and that the image of x is the surface 
M.z = (x 2 - y 2 )/4:. 
10. If F is an isometry of E 3 and M is a surface in E 3 , prove that the image 



Sec. 2] PATCH COMPUTATIONS 133 




FIG. 4.11 

F(M) is also a surface in E 3 . (Hint: If x is a patch in M, then the 
composite function F(x) is regular, since F(x)* = F* x* by Ex. 12 
of 1.7.) 

11. The assertion in Exercise 10 remains true when F is merely a diffeo- 
morphism. Prove this special case: If F is a diffeomorphism of E 3 , then 
the image of the surface M: g = c is M: g = c, where g = g(F~ ) — 
and M is a surface. (Hint: If dg(v) 9* at pin M , show by using Ex. 9 
of 1.7 that dg (F* v) 5* 0. 

12. If / is a differentiable function and f(x,y 2 ) = c defines a curve C in 
the xy plane, then C is symmetric about the x axis and must cross this 
axis once (if C is an arc) or twice (if C is closed). Prove that revolving 
C about the x axis gives a surface M in E 3 . We call M an augmented 
surface of revolution: If the points on the axis are deleted, it becomes 
an ordinary surface of revolution (Fig. 4.11). 



2 Patch Computations 

In Section 1, coordinate patches were used to define a surface; now we con- 
sider some properties of patches that will be useful in studying surfaces. 

Let x: D — » E 3 be a coordinate patch. Holding u or v constant in the 
function (w, v) — * x(w, v) produces curves. Explicitly, for each point 
(uo, vo) in D the curve 

u —*■ x(u, Vo) 

is called the u-parameter curve, v = Vo, of x; and the curve 

v — > x(u , v) 

is the v-parameter curve, u = u (Fig. 4.12). 

Thus the image x(D) is covered by these two families of curves, which 
are the images under x of the horizontal and vertical lines in D, and one 
curve from each family goes through each point of x(D). 



134 



CALCULUS ON A SURFACE 



[Chap. IV 



V 



(Wo , V ) 



u-paxameter curves 

u = Mo. 



D 




( M-parameter 
curves 



E a 



u 



x(mo , v ) 



V = v 



FIG. 4.12 



x«(mo , «o) 



x.(wo , t> ) 




2.1 Definition If x: D -> E 3 is 



x(t*o , i>o) 



FIG. 4.13 



a patch, for each point (wo, v ) in 
D: 

(1) The velocity vector at Uq 
of the w-parameter curve, v = v , 
is denoted by x u (uo, v ). 

(2) The velocity vector at y 
of the y-parameter curve, u = Uo, 
is denoted by x v (u , v ). 

The vectors x u (uo, v ) and x v (u , v ) are called the partial velocities of x 
at (uo,v ) (Fig. 4.13). 

Thus x„ and x„ are actually functions on D whose values at each point 
(u , v ) are tangent vectors to E at x(u , v ). The subscripts u and v are 
intended to suggest partial differentiation. Indeed if the patch is given in 
terms of its Euclidean coordinate functions by a formula 

x(u,v) = (xi(u,v),x 2 (u,v),X3(u,v)), 

then it follows from the definition above that the partial velocity functions 
are given by 



(dx\ dx 2 dxz\ 

du ' du du Jx 

(dxi dXi dxs\ 

dv ' dv dv/x 



The subscript x (frequently omitted ) is a reminder that x u (u, v) and x„ (u, v) 
have point of application x(u, v). 

2.2 Example The geographical patch in the sphere. Let 2 be the sphere 
of radius r > centered at the origin of E . Longitude and latitude on the 
earth suggest a patch in S quite different from the Monge patch used on 



Sec. 2] 



PATCH COMPUTATIONS 



135 



D 



<u,v) 




FIG. 4.U 



S in Section 1. The point x(u, v) of 2 with longitude u ( — -k < u < ir) 
and latitude v ( — x/2 < v < ir/ 2 ) has Euclidean coordinates (Fig. 4.14). 

x(u, v) = (r cos v cos w, r cos y sin w, r sin v). 

With the domain D of x denned by these inequalities, the image x(D) 
of x is all of 2 except one semicircle from north pole to south pole. The 
w-parameter curve, v = vo, is a circle — the parallel of latitude vo. The 
v-parameter curve, u = u , is a semicircle — the meridian of longitude uo. 

We compute the partial velocities of x to be 

x u (u, v) = r( — cos v sin u, cos v cos u, 0) 

x^w, v) = r( — sin v cos u, —sin v sin w, cos i>) 

where r denotes a scalar multiplication. Evidently x u always points due 
east, and x„ due north. In a moment we shall give a formal proof that x is 
a patch in 2 (Fig. 4.15). 

To test whether a given subset M of E 3 is a surface, Definition 1.2 de- 
mands proper patches (and Example 1.7 shows why). But once we know 
that M is a surface, the properness condition need no longer concern us 
(Exercise 14 of Section 3). Furthermore, in many situations the one-to-one 
restriction on patches can also be dropped. 

2.3 Definition A regular mapping x: D — » E 3 whose image lies in a surface 
M is called a parametrization of the region x(Z>) in M. 

(Thus a patch is merely a one-to-one parametrization.) In favorable 
cases this image x(D) may be the whole surface M, and we have then the 
analogue of the more familiar notion of parametrization of a curve (p. 20). 
Parametrizations will be of first importance in practical computations with 



136 



CALCULUS ON A SURFACE 



[Chap. IV 




x„(Wo , »o) 



meridian 

u — Mo 



FIG. 4.15 

surfaces, so we consider some ways of determining whether a mapping 
x: D — ► E is a parametrization of (part of ) a given surface M. 

The image of x must, of course, lie in M. Note that if the surface is given 
in the implicit form M: g = c, this means that the composite function g(x) 
must have constant value c. 

To test whether x is regular, note first that parameter curves and partial 
velocities x„ and x„ are well-defined for an arbitrary differentiable mapping 
x: D — > E . Now the last two rows of the cross product 

*7i U* U s 
dxi 
x u X x„ = du 
dxi 
~dv 

give the (transposed) Jacobian matrix of x at each point. Thus the regu- 
larity of x is equivalent to the condition that x M X x„ is never zero, or, by 
properties of the cross product, that at each point (u, v) of D the partial ve- 
locity vectors of x are linearly independent. 

Let us try out these methods on the mapping x given in Example 2.2. 
Since the sphere is defined implicitly by g = x 2 + y 2 + z 2 = r 2 , we must 
show that g (x) = r 2 . Substituting the coordinate functions of x for x, y, and 
z, we get 

r~ 2 g(x) = (cos v cos u) 2 -f (cos v sin u) 2 + sin 2 u 

= cos 2 y + sin 2 ?; = 1. 

A short computation using the formulas for x u and x„ given in Example 
2.2 yields 

r~ 2 x„ X x„ = cos u cos 2 y U\ + sin u cos 2 y £/ 2 + cos v sin v U 3 . 



dx 2 
du 


dx 3 
du 


dx% 

dv 


dxz 
dv 



See. 2] 



PATCH COMPUTATIONS 



137 



Since -tt/2 < v < x/2 for the domain D of x, cos v is never zero there; 
but sin u and cos u are never simultaneously zero, so x u X x„ is never zero 
on D. Thus x is regular, and hence is a parameterization. 

To show that x is a patch, we prove it is one-to-one, that is, show that 

x(w, v) = x(wi, vi) implies (u, v) = (wi, Vi). If x(w, y) = x(ui, Vi), then 
the definition of x gives three coordinate equations: 

r cos v cos u — r cos Wi cos U\ 

r cos v sin u = r cos Vi sin Wi 
r sin v = r sin w x . 

Again, since -w/2 < v < tt/2 for all points of D, the last equation 
implies that v = Vi. Thus r cos v = r cos t>i > may be cancelled from the 
first two equations and we conclude that u = U\ also. 

For this particular function x in 2, these results might almost be con- 
sidered obvious from the discussion in Example 2.2, but the methods used 
above will serve in more difficult cases. 

We shall now see how to find natural parametrizations in cylinders and 
surfaces of revolution. 

2.4 Example Parametrization of a cylinder M. Suppose, as in Ex- 
ample 1.5, that M is the cylinder over a curve C: f(z, y) = a in the xy 
plane (Fig. 4.16). If a = («i, a 2 , 0) is a parametrization of C, we assert 
that 

x(u,v) = («i(w), a 2 (u),v) 




(ai(u), ai(w), 0) 
FIG. 4.16 



138 



CALCULUS ON A SURFACE 



[Chap. IV 



is a parametrization of M . Clearly x lies in M, covers all of M , and is differ- 
entiable. Furthermore x is regular, for at each point (u, v) the partial 
velocities 



_ /dai daz \ 
Xu " \du ' du ' V 

x, = (0, 0, 1) 



are linearly independent (x u is never zero, since a is by definition regular). 
If the curve a is defined on an interval J, the domain of x is the vertical 
strip D: u in /, v arbitrary. (Thus if / is the whole real line, D is E 2 .) The 
w-parameter curves of x are merely translates of C and are called the 
cross-sectional curves of the cylinder. The v-parameter curves follow the 
straight lines called the rulings (or "elements") of the cylinder. If C is not 
a closed curve, then a — and hence also x— is one-to-one, so x is a patch. 
But if C is closed, x wraps D an infinite number of times around C. 

2.5 Example Parametrization of a surface of revolution. Suppose that 
M is obtained, as in Example 1.6, by revolving a curve C in the upper half 
of the xy plane about the x axis. Now let 

a(u) = (g(u),h(u), 0) 

be a parametrization of C (note that h > 0). As we observed in Example 
1.6, when the point (g(u), h(u), 0) on the profile curve C has been rotated 
through an angle v, it reaches a point x(u, v) with the same x coordinate 
g(u), but new y and z coordinates h(u) cos v and h(u) sin v, respectively 
(Fig. 4.17). Thus 

x(u, v) = (g(u), h(u) cos v, h(u) sin v) 



(jf(u), *(tt), 0) 

Z 




FIG. 4.17 



Sec. 2] 



PATCH COMPUTATIONS 



139 



Evidently this formula defines a mapping into M whose image is all of M. 
A short computation shows that x„ and x v are always linearly independent, 
so x is a parametrization of M. As in Example 2.4, the domain D of x con- 
sists of all points (u, v) for which u is in the domain of a. The w-parameter 
curves of x parametrize the meridians of M; the v-parameter curves, the 
parallels. (Thus the parametrization x: D —*■ M is never one-to-one.) 

Obviously we are not limited to rotating curves in the xy plane about 
the x axis. But with other choices of coordinates, we maintain the same 
geometric meaning for the functions g and h: g measures distance along 
the axis of revolution, while h measures distance from the axis of revolu- 
tion. 

Actually the geographical patch in the sphere is one instance of Example 
2.5 (with u and v reversed) ; here is another. 

2.6 Example Torus of revolution T. This is the surface of revolution 
obtained when the profile curve C is a circle. Suppose that C is the circle 
in the xz plane with radius r > and center (R, 0, 0). We shall rotate 
about the z axis; hence we must require R > r to keep C from meeting 
the axis of revolution. A natural parametrization (Fig. 4.18) for C is 

a(u) = (R -\- r cos u, r sin u). 

Thus by the remarks above we must have g(u) = r sin u, distance along 
the z axis, and h(u) = R + r cos u, distance from the z axis. The general 
argument in Example 2.5 — with coordinate axes permuted — then yields the 
parametrization 

x(u, v) = (h(u) cos v, h(u) sin v, g{u)) 

= ((R -\- r cos u) cos v, (R + r cos u) sin v, r sin u). 
The domain of x is the whole plane E 2 , and (as always when the profile 
curve is closed) x is periodic in both u and v. Here 

x(m + 2ir, v + 2x) = x(w, v) for all (u, v). 
There are infinitely many different parametrizations (and patches) in 




FIG 4.18 



FIG. 4.19 



140 CALCULUS ON A SURFACE [Chap. IV 

any surface. Those we have discussed have been singled out by the natural 
way they are fitted to their surfaces. 



EXERCISES 

1. Find a parametrization for the entire surface obtained by revolving: 

(a) C: y = cosh x around the x axis (catenoid). 

(b) C: (z — 2) 2 + y 2 = 1 around the y axis (torus). 

(c) C: z = x 2 around the z axis (paraboloid of revolution). 

2. Partial velocities x M and x„ are defined for an arbitrary mapping x: 
D — > E , so we may consider the real-valued functions 

E = x u »x u F = x u «x„ G = x v *x v 

on D. Prove 

|| x u X x v || 2 = EG- F 2 . 

Deduce that x is a regular mapping if and only if EG — F 2 is never 
zero. (This is often the easiest way to check regularity. The geometric 
significance of these functions is discussed in V.4). 

3. Show that 

M: (y/ x * + y 2 - 4) 2 + z = 4 

is a torus of revolution: Find a profile circle and the axis of revolution. 

A ruled surface is a surface swept out by a straight line L moving 
along a curve 0. The various positions of the generating line L are 
called the rulings of the surface. Such a surface thus always has a 
parametrization in ruled form 

x(u,v)=p(u) + v8(u) or @(v) + u8(v) 

where we call /J the base curve, 8 the director curve. Alternatively we 
may vizualize 8 as a vector field on /?. Frequently it is necessary to 
restrict v to some interval, so the rulings may not be entire straight 
lines. 

4. Show that the saddle surface M: z = xy is doubly ruled: Find two 
ruled parametrizations with different rulings. 

5. A cone is a ruled surface with parametrization of the form 

x(u, v) = p + v 8(u). 

Thus all rulings pass through the vertex p (Fig. 4.20). Show that the 
regularity of x is equivalent to both v and 5X5' never zero. (Thus the 
vertex is never part of the cone. ) 



Sec. 2] 



PATCH COMPUTATIONS 



141 





FIG. 4.20 



FIG. 4.21 



6. A cylinder is a ruled surface with parametrization of the form 

x(u, v) = |8(w) + vq. 

Thus the rulings are all parallel (Fig. 4.21). Prove that the regularity 
of x is equivalent to £ X q never zero. Show that this definition 
generalizes Example 2.4. 

7. A line L is attached orthogonally to an axis A (Fig. 4.22 ). If L moves 
along A and rotates — both at constant speed — then L sweeps out a 
helicoid H. 

If A is the z axis, then H is the image of the mapping x: E 2 — > E 3 such 
that 

x(u, v) = (u cos v, u sin v, bv) 

(b * 0). 

(a) Prove that x is a patch. 

(b ) Describe the parameter curves 
of x. 

(c ) Express the helicoid in implicit 
form g = c. 

8. (a) Show that x: D ,— > E 3 is a reg- 

ular mapping, where 

\(u, v) = (u cos v, u, sin v) 

onD:u>0. /^ (m cos v, u sin t>, 0) 

(b) Find a function g(x, y, z) such 
that the image of x is the sur- fig. 4.22 
face M : g = 0. 

(c) Show that M is a ruled surface and sketch M. (Hint: Begin with 
the curve sliced from M by the plane y = 1.) 

9. Let jS be a unit-speed parametrization of the unit circle in the xy plane. 



(0,0, 





142 CALCULUS ON A SURFACE [Chap. IV 

Construct a ruled surface as follows: Move a line L along in such a 
way that L is always orthogonal to the radius of the circle and makes 
constant angle 7r/4 with &' (Fig. 4.23). 

(a) Derive this parametrization of the resulting ruled surface M: 

x(u,v) = 0(u) + v(p'(u) + Uz). 

(b) Express x explicitly in terms of v 
and coordinate functions for 0. 

(c) Deduce that M is given implicitly 
by the equation 

x 2 + y 2 - i = 1. 

(d) Show that if the angle 7r/4 above 

is changed to — t/4, the same surface M results. Thus M is 
doubly ruled. 

(e) Sketch this surface M showing the two rulings through each of 
the points (1, 0, 0) and (2, 1, 2). 

A quadrie surface is a surface M: g = for which g involves at most 
quadratic terms in x x , x 2 , x 3 — that is, 

9 = iLi anXiXj + 2 Mi + c. 

i,j i 

Trivial cases excepted, there are just five types of quadrie surfaces, repre- 
sented by the familiar surfaces in the next three examples (see Theorem 
2.2, p. 280, Birkhoff and MacLane [2]). 

10. In each case, (i) prove that M is a surface and sketch its general 
shape, (ii) show that x is a parametrization and find its image in M. 

(a) Ellipsoid, M:- 2 + l- + Z - = l 

a 2 b 2 c 2 

x(u, v) = (a cos u cos v, b cos u sin v, c sin u) 

on/): -x/2 < u < x/2. 

2 2 2 

X 1J z 

(b) Elliptic hyperboloid, M: — + ^- = 1 

a 2 b 2 c 2 

x(u, v) = (a cosh u cos v, b cosh u sin v, c sinh u) on E 2 . 

2 2 2 

(c) Elliptic hyperboloid (two sheets), M: (- — = —1 

a 2 b 2 c 2 

x (u, v) = (a sinh u cos v, b sinh u sin v, c cosh u) onD.u^O. 

2 2 

X 1J 

11. Elliptic paraboloid, M: z = — + f- . 

a 2 o 2 

(a) Show that M is a surface, and that 



Sec. 3] DIFFERENTIABLE FUNCTIONS AND TANGENT VECTORS 143 

x(u, v) = (au cos v, bu sin v, w 2 ), u > 0, 

is a parametrization that omits only one point of M. 
(b) Describe the parameter curves of x in general, and sketch this 
surface for a = 1, b = 4, showing some parameter curves. 

2 2 

X V 

12. Hyperbolic paraboloid, M: z = - - — fr- . 

a 2 cr 

(a) Show that x: E 2 — > E is a proper patch covering all of M, where 

x(u, v) = (a(u -f v), b(u — v), 4uv). 

(b) Show that M is a doubly ruled surface by rewriting x in ruled 
form in two different ways. 

(c) Same as (b) of Exercise 11. 

13. Let M be the surface of revolution obtained by revolving the curve 
t — > (g(t), h(t), 0) about the x axis (h > 0). Show that 

(a) If gr is never zero, M has a parametrization of form 

x(w, v) = (u, f(u) cos v, f(u) sin v). 

(b) If h is never zero, M has a parametrization of form 

x(w, v) = (f(u), u cos v, u sin v). 



3 Differentia ble Functions and Tangent Vectors 

We now begin an exposition of the calculus on a surface M in E 3 . The 
space E will gradually fade out of the picture, since our ultimate goal is a 
calculus for M alone. Generally speaking, we shall follow the order of topics 
in Chapter I, making such changes as are necessary to adapt the calculus 
of the plane E 2 to a surface M. 

Suppose that / is a real-valued function denned only on a surface M. 
If x: D — > M is a coordinate patch in M, then the composite function 
/(x) is called a coordinate expression for /; it is an ordinary real -valued 
function (u,v) — > /(x (u, v) ). We define / to be differentiable provided all its 
coordinate expressions are differentiable in the usual Euclidean sense 
(Definition 1.3 of Chapter I). 

For a function F: E n — > M, each patch x in M gives a coordinate expres- 
sion x~ (F) for F. Evidently this composite function is defined only on the 
set G of all points p of E n such that F(p) is in x(D). Again we define F 
to be differentiable provided all its coordinate expressions are differentiable 
in the usual Euclidean sense. [We must understand that this includes the 



144 



CALCULUS ON A SURFACE 



[Chap. IV 



requirement that e be an open set of E n , so that the differentiability of 
x -1 (F) : — >• E 2 is well-defined, as in Section 7 of Chapter I.] 

In particular, a curve a: I — > M in a surface M is, as before, a differentia- 
ble function from an open interval / into M. 

To see how this definition works out in practice, we examine an important 
special case. 

3.1 Lemma If a is a curve a: I —*■ M whose route lies in the image x(D) 
of a single patch x, then there exist unique differentiate functions Oi, ch 
on / such that 

a(t) = x(ai(t), a 2 (t)) for all I 

or, in functional notation, a = x(a u a 2 ). (See Fig. 4.24.) 

Proof. By definition, the coordinate expression x~ x a: J — *■ D is differentia- 
ble — it is just a curve in E 2 whose route lies in the domain D of x. If a h az 
are the Euclidean coordinate functions of x -1 a, then 

a = xx a = x(oi, 02). 

These are the only such functions, for if a = x(6i, 6 2 ), then 

(«i, 02) = x _1 a = x _1 x(6i, 6 2 ) = (61, &2>. I 

These functions ai, a- 2 are called the coordinate functions of the curve a 
with respect to the patch x. For example, the curve a given in (3 ) of Exam- 
ple 4.2, Chapter I, lies in the part of the sphere 2 of radius 2 that is covered 
by the patch x given in Example 2.2. Observe that this curve moves so as 
to have equal longitude and latitude at each point. In fact its coordinate 
functions with respect to x are ai(t) = (h(t) = t, since by the formula 
for x, 

x(a,i(t), (h(t)) = x(t, t) = 2(cos 2 f, cos t sin t, sin t) = a(t). 

For an arbitrary patch x: D — > M (as in the case just considered) it is 



<x(t)= x( ai (t), 0,(0) 




FIG. 4.24 



Sec. 3] 



DIFFERENTIABLE FUNCTIONS AND TANGENT VECTORS 



145 



natural to think of the domain D as a map of the region x(D) in M. The 
functions x and x _1 establish a one-to-one correspondence between objects 
in x(Z>) and objects in D. If a curve a in x(D) represents the voyage of a 
ship, the coordinate curve (ai, ck) plots its position on the map D. 

A rigorous proof of the following rather technical fact requires the meth- 
ods of advanced calculus, and we shall not attempt to give a proof here. 

3.2 Theorem Let M be a surface in E 3 . If F : E n — > E 3 is a (diff erentiable ) 
mapping whose image lies in M, then considered as a function F: E n — ► M 
into M, F is differentiable (as on p. 143 . ) 

This theorem links the calculus of M tightly to the calculus of E 3 . For 
example, it implies the "obvious" result that a curve in E 3 which lies in 
Mis a curve of M. 

Since a patch is a differentiable function from (an open set of) E into 
E 3 , it follows that a patch is a differentiable function into M . Hence its 
coordinate expressions are all differentiable, so patches overlap smoothly. 

3.3 Corollary If x and y are patches in a surface M in E whose images 
overlap, then the composite functions x _1 y and y _1 x are (differentiable) 
mappings defined on open sets of E 2 . 

(The function y -1 x, for example, is defined only for those points (u, v) 
in D such that x(u,v) lies in the image y(E) of y (Fig. 4.25). 
By an argument like that for Lemma 3.1, Corollary 3.3 can be rewritten. 

3.4 Corollary If x and y are overlapping patches in M, then there exist 
unique differentiable functions u and v such that 

y(u,v) = x(u(u,v),v(u,v)) 

for all (u, v) in the domain of x -1 y. In functional notation: y = x(w, v). 

There are, of course, symmetrical equations expressing x in terms of y. 
Corollary 3.3 makes it much easier to prove differentiability. For exam- 
ple, if / is a real- valued function on M, instead of verifying that all coordi- 




146 CALCULUS ON A SURFACE [Chap. IV 

nate expressions /(x) are Euclidean differentiable, we need only do so for 
enough patches x to cover all of M (so a single patch will often be enough). 
The proof is an exercise in checking domains of composite functions: 
For an arbitrary patch y, fx and x _1 y differentiable imply /xx -1 y differen- 
tiable. This function is in general not /y, because its domain is too small. 
But since there are enough x's to cover M , such functions constitute all 
of /y, and thus prove it is differentiable. 

It is intuitively clear what it means for a vector to be tangent to a 
surface M in E 3 . A formal definition can be based on the idea that a curve in 
M must have all its velocity vectors tangent to M. 

3.5 Definition Let p be a point of a surface M in E 3 . A tangent vector 
v to E at p is tangent to M at p provided v is a velocity vector of some 
curve in M (Fig. 4.26). 

The set of all tangent vectors to M at p is called the tangent plane of M 
at p, and is denoted by T P (M). The following result shows, in particular, 
that at each point p of M the tangent plane T p (M) is actually a two-dimen- 
sional vector subspace of the tangent space T P (E 3 ). 

3.6 Lemma Let p be a point of a surface M in E 3 , and let x be a patch 
in M such that x(w , v ) = p. A tangent vector v to E 3 at p is tangent 
to M if and only if v can be written as a linear combination of x u (u Q , v ) 
and x v (u , v ). 

Since partial velocities are always linearly independent, we deduce that 
they provided a basis for the tangent plane of M at each point of x(D). 

Proof. Note that the parameter curves of x are curves in M, so these 
partial velocities are tangent to M at p. 

First suppose that v is tangent to M at p; thus there is a curve a in M 
such that a(0) = p and a (0) = v. Now by Lemma 3.1, a may be written 
a = x(ai, a 2 ); hence by the chain rule 



T V (M) 




FIG. 4.26 



Sec. 3] DIFFERENTIABLE FUNCTIONS AND TANGENT VECTORS 147 

/ / \ da\ , N da,2 

a = x„Ui, a 2 J -j- + x„(ai, a 2 ; -=- • 
at at 

But since a(0) = p = x(w , v ), we have Oi(0) = u , a 2 (0) = i>o. Hence 
evaluation at t = yields 

v = a'(0) = -r- 1 (0) x„(w , Wo) + -rr (0) x "( w o, Wo). 
at at 

Conversely, suppose that a tangent vector v to E 3 can be written 

v = Cix u (u , v ) + c 2 x„(w , v ). 

By computations as above, v is the velocity vector at t — of the curve 

t — > x(w + tci, v Q + <c 2 ). 

Thus v is tangent to M at p. | 

A reasonable deduction, based on the general properties of derivatives, 
is that the tangent plane T P {M) is the linear approximation of the surface 
M near p. 

3.7 Definition A Euclidean vector field Z on a surface M in E 3 is a func- 
tion that assigns to each point p of M a tangent vector Z(p) to E 3 at p. 

A Euclidean vector field V for which each vector V(p) is tangent to M 
atp is called a tangent vector field on M (Fig. 4.27). Frequently these vector 
fields are defined, not on all of M, but only on some region in M . As usual, 
we always assume differentiability (Exercise 12). 

A Euclidean vector z at a point p of M is normal to M if it is orthogonal 
to the tangent plane T P {M) — that is, to every tangent vector to M at p. 
And a Euclidean vector field Z on M is a normal vector field on M provided 
each vector Z(p) is normal to M. 

Because T P (M) is a two-dimensional subspace of ^(E ), there is only 
one direction normal to M at p: All normal vectors z at p are collinear. 

Z(p)> 




FIG. 4.27 



148 CALCULUS ON A SURFACE [Chap. IV 

Thus if z is not zero, it follows that T P (M) consists of precisely those vectors 
in T P (F, ) that are orthogonal to z. 

It is particularly easy to deal with tangent and normal vector fields on a 
surface given in implicit form. 

3.8 Lemma If M: g = c is a surface in E 3 , then the gradient vector field 
Vg = X- (dg/dXi)Ui (considered only at points of M ) is a non vanishing 
normal vector field on the entire surface M . 

Proof. The gradient is nonvanishing (that is, never zero) on M since by 
Theorem 1.4 the partial derivatives dg/dxi cannot simultaneously be 
zero at any point of M. 

We must show that (V</)(p)»v = for every tangent vector v to M 
at p. First note that if a is a curve in M, then g(a) = g(ai, a 2 , 0:3) has 
constant value c. Thus by the chain rule 

dXi at 

Now choose a to have initial velocity 

a'(0) = v = (v u V 2 , V 3 ) 
ata(O) = p. Then 

°" Eg wo)) f (0) - £g(p)* " (V9)(pW " a ■ 

3.9 Example Vector fields on the sphere 2 : g = 2x, 2 = r 2 . The lemma 
shows that 

X = \ Vg = S XiUi 

is a normal vector field on S (Fig. 4.28). This is geometrically evident, since 
X(p) = 2 PiUi(p) is the vector p with point of application p! It follows 
by a remark above that \ p is tangent to 2 if and only if the dot product 



*(P) = Pp 




FIG. 4.28 



Sec. 3] DIFFERENTIABLE FUNCTIONS AND TANGENT VECTORS 149 

y p'Pp — v«p is zero. Similarly a vector field V on 2 is a tangent vector 
field if and only if V'X = 0. For example, F(p) = (— p 2 , pi, 0) defines 
a tangent vector field on 2 that points "due east" and vanishes at the north 
and south poles (0, 0, ±r). 

We must emphasize that only the tangent vector fields on M belong to 
the calculus of M itself, since they derive ultimately from curves in M 
(Definition 3.5). This is certainly not the case with normal vector fields. 
However, as we shall see in the next chapter, normal vector fields are quite 
useful in studying M from the viewpoint of an observer in E 3 . 

Finally, we shall adapt the notion of directional derivative to a surface. 
Definition 3.1 of Chapter I uses straight lines in E ; thus we must use the 
less restrictive formulation based on Lemma 4.6 of Chapter I. 

3.10 Definition Let v be a tangent vector to M at p, and let / be a differ- 
entiate real-valued function on M. The derivative v[/] of f with respect to v 
is the common value of (d/dt)(fa) (0) for all curves a in M which have 
initial velocity v. 

Directional derivatives on a surface have exactly the same linear and 
Leibnizian properties as in the Euclidean case (Theorem 3.3 of Chapter I) . 



EXERCISES 

1. Let x be the geographical patch in the sphere 2 (Ex. 2.2). Find the 
coordinate expression /(x) for the following functions on 2: 

(a) /(p) = Pl 2 + p 2 2 , (b) f(p) = ( Pl - p 2 ) 2 + Pz- 

2. Let x be the parametrization of the torus given in Example 2.6. 

(a) Find the Euclidean coordinates ai, a 2 , a 3 of the curve a (t) = x (t, t). 

(b) Show that a is periodic, and find its period (see p. 20). 

3. (a) Prove Corollary 3.4. 

(b) Derive the "chain rule" 

_ du dv _ du dv 

Yu ~ du Xu + ^ Xv Yv ~ dv~ Xu + dv Xv 
where x„ and x„ are evaluated on (u,v). 

(c) Deduce that y„ X y, = Jx u X x„, where J is the Jacobian of the 
mapping x -1 y = (u,v): D — > E 2 . 

4. Let x be a patch in M . 

(a) If x* is the derivative map of x (1.7), show that 

x*(£/i) = x u x*(£/ 2 ) = x„ 
where Ui, U 2 is the natural frame field on E 2 . 



150 CALCULUS ON A SURFACE [Chap. IV 

(b) If / is a differentiable function on M , prove 

xJ/] = ^(/(x)) x v [f] =!;(/(*)). 

5. Prove that : 

(a) v is tangent to M: z - f(x, y) at a point p of M if and only if 

Vz = dx ^ Ph v ^ Vl + d ^ Ph V ^ V2 ' 

(b) if x is a patch in an arbitrary surface M, then v is tangent to M 
at x(u, v) if and only if 

vx„(w, v) X x„(i*, v) = 0. 

6. Let x and y be the patches in the unit sphere 2 that are defined on the 
unit disc D: u 2 + v < 1 by 

x(u,v) = (u,v,f(u,v)) y(u,v) = (v,f(u,v),u) 

where/ = \/l — u 2 — v 2 . 

(a) On a sketch of 2 indicate the images x(D) and y(D), and the 
region on which they overlap. 

(b) At which points of D is y -1 x defined? Find a formula for this 
function. 

(c) At which points of D is x _1 y defined? Find a formula for this 
function. 

7. Find a nonvanishing normal vector field onM: z = xy and two tangent 
vector fields that are linearly independent at each point. 

8. Let C be the right circular cone parametrized by 

x(u, v) = y(cos u, sin u, 1). 

If a is the curve a(t) = x(-\/2t, e l ) 

(a) Express a in terms of x„ and x„. 

(b) Show that at each point of a, the velocity a bisects the angle 
between x M and x„. {Hint: Verify that a *x u /\\ x u || = a •x„/|| x„ ||, 
where Xu and x» are evaluated on (y/2t, e 1 ).) 

(c) Make a sketch of the cone C showing the curve a. 

9. If z is a nonzero vector normal to M at p, let T P (M) be the plane 
through p orthogonal to z (see page 60). Prove: 

(a) If each tangent vector \ p to M at p is replaced by its tip p + v, 
then T P (M) becomes T P (M). (Thus f p {M) gives a concrete 
representation of T p (M ) in E 3 . ) 

(b) If x is a patch in M, then f\( U ,v)(M) consists of all points r in E 
such that (r — x(u,v))*x u (u, v) X x v (u, v) = 0. 



Sec. 3] DIFFERENTIABLE FUNCTIONS AND TANGENT VECTORS 151 

(c) If M is given implicitly by g = c, then f p (M) consists of all 
points r in E 3 such that (r — p) • (Vg) (p) = 0. 

10. In each case find an equation of the form ax + by + cz = d for the 
plane T P (M). 

(a) p = (0, 0, 0) and M is the sphere 

x 2 + y 2 + (z- 1) 2 = 1. 

(b) p = (1, —2, 3) and M is the ellipsoid 

2 2 2 

4 ^ 16 ^ 18 

(c) p = x(2, 7r/4), where M is the helicoid parametrized by 

x(u, v) = (u cos v, u sin v, 2v). 

1 1 . (Continuation of Exercise 2 ) . 

(a) If m and n are integers with greatest common divisor 1, show that 
a(t) = x(mt, nt) is a simple closed curve on the torus, and find 
its period. f 

(b) If q is an irrational number, show that the curve a: R — >• T such 
that <x(t) = x(t, qt) is one-to-one. 

This curve, called a winding line on the torus T, is dense in 
T; that is, given any number e > 0, a comes within distance e 
of every point of T. 

12. A Euclidean vector field Z = J] z i Ui on M is differentiable provided 
its coordinate functions z u z 2 , z 3 (on M) are differentiable. If V is a 
tangent vector field, show that 

(a) For every patch x: D — > M, V can be written as 

V(x(u,v)) = f(u,v)x u (u,v) + g(u,v)x v (u,v) 

(b) V is differentiable if and only if the functions / and g (on D) are 
differentiable. 

The following exercises require some knowledge of point-set topology. 
They deal with open sets in a surface M in E 3 , that is, sets ai in M which 
contain a neighborhood in M of each of their points. 

13. Prove that if y: E — -> M is a proper patch, then y carries open sets in 
E to open sets in M. Deduce that if x: D -> M is an arbitrary patch, 
then the image x(D) is an open set in M. (Hint: To prove the latter 
assertion, use Corollary 3.3.) 

t That is, show a(t') = a(t) if and only if t' — t is a multiple of the period p. 
Roughly speaking, this means the route of a is O-shaped rather than, say, 8-shaped. 



152 CALCULUS ON A SURFACE [Chap. IV 

14. Prove that every patch x: D — » M in a surface M in E 3 is proper. 
(Hint: Use Exercise 13. Note that (x _1 y)y _1 is continuous and agrees 
with x _1 on an open set in x(D).) 

15. If 01 is a subset of a surface M in E 3 , prove that 01 is itself a surface 
in E if and only if 01 is an open set of M . 



4 Differential Forms on a Surface 

In Chapter I we discussed differential forms on E 3 only in sufficient detail 
to take care of the Cartan structural equations (Theorem 8.3 of Chapter 
II). In the next three sections we shall give a rather complete treatment of 
forms on a surface. 

Forms are just what we need to describe the geometry of a surface 
(Chapters VI and VII), but this is only one example of their usefulness. 
Surfaces and Euclidean spaces are merely special cases of the general notion 
of manifold (Section 8). Every manifold has a differential and integral 
calculus — expressed in terms of forms — which generalizes the usual ele- 
mentary calculus on the real line. Thus forms are fundamental to all the 
many branches of mathematics and its applications that are based on cal- 
culus. In the special case of a surface, the calculus of forms is rather easy, 
but still gives a remarkably accurate picture of the most general case. 

Just as for E 3 , a 0-formf on a surface M is simply a (differentiable) real- 
valued function on M, and a 1-form <f> on M is a real- valued function on 
tangent vectors to M that is linear at each point (Definition 5.1 of Chapter 
I). We did not give a precise definition of 2-forms in Chapter I, but we shall 
do so now. A 2-formwillbe a two-dimensional analogue of a 1-form: a real- 
valued function, not on single tangent vectors, but on pairs of tangent 
vectors. (In this context the term "pair" will always imply that the tangent 
vectors have the same point of application.) 

4.1 Definition A 2-form y\ on a surface M is a real-valued function on all 
ordered pairs of tangent vectors v,w to I such that 

(1 ) t] (v, w) is linear in v and in w. 

(2) r/(v,w) = -t)(w,v). 

Since a surface is two-dimensional, all p forms with p > 2 are zero, by 
definition. This fact considerably simplifies the theory of differential forms 
on a surface. 

At the end of this section we will show that our new definitions are con- 
sistent with the informal exposition given in Chapter I, Section 6. 

Forms are added in the usual pointwise fashion; we add only forms of 
the same degree p = 0, 1, 2. Just as we evaluated a 1-form <j> on a vector 



Sec. 4] DIFFERENTIAL FORMS ON A SURFACE 153 

field V, we now evaluate a 2-form ijona pair of vector fields V, W to get a 
real- valued function ri(V,W) on the surface M. Of course we shall always 
assume that the forms we deal with are differentiable — that is, convert 
(differentiable) vector fields into differentiable functions. 
Note that the alternation rule (2) in Definition 4.1 implies that 

v(v,\) = 

for any tangent vector v. This rule also shows that 2-forms are related to 
determinants. 

4.2 Lemma Let if be a 2-form on a surface M, and let v and w be (line- 
arly independent) tangent vectors at some point of M. Then 



?/(av + bw, cv + dw) = 



a b 
c d 



v(y, w) 



Proof. Since 17 is linear in its first variable, its value on the pair of tangent 
vectors av + bw, cv + dw is av(\, cv + dw) + brj (w, cv + dw). Using 
the linearity of i\ in its second variable, we get 

ac t?(v,v) + adrj(\,w) + be r?(w,v) + bdrf(w,w). 

Then the alternation rule (2 ) gives 

77 (av + bw, cv + dw) = (ad — be) ??(v,w). | 

Thus the values of a 2-form on all pairs of tangent vectors at a point 
are completely determined by its value on any one linearly independent pair. 
This remark is used frequently in later work. 

Wherever they appear, differential forms satisfy certain general proper- 
ties, established (at least partially) in Chapter I for forms on E 3 . To begin 
with, the wedge -product of a p-form and a q-form is always a (p + q)-form. 
If p or q is zero, the wedge product is just the usual multiplication by a 
function. On a surface, the wedge product is always zero if p + q > 2. So 
we need a definition only for the case p = q = 1. 

4.3 Definition If <f> and \p are 1-forms on a surface M, the wedge product 
4> a \p is the 2-form on M such that 

(<t> a ^)(v, w) = 0(v) ^(w) - <f>(w) rf/(\) 
for all pairs v, w of tangent vectors to M. 

Note that <f> a if/ really is a 2-form on M, since it is a real- valued function 
on all pairs of tangent vectors and satisfies the conditions in Definition 4.1. 
The wedge product has all the usual algebraic properties except commuta- 
tivity; in general, if £ is a p-form and 17 is a q-form, then 

€ A „ = (-!)%, A $. 



154 CALCULUS ON A SURFACE [Chap. IV 

On a surface the only minus sign occurs in the multiplication of 1 -forms, 
where just as in Chapter I, we have <£ a \f/ = — \f> a <j>. 

The differential calculus of forms is based on the exterior derivative d. 
For a 0-form (function) / on a surface, the exterior derivative is, as before, 
the 1-form df such that df(\) = vf/]. Wherever forms appear, the exterior 
derivative of a p-form is a (p + l)-form. Thus for surface the only new 
definition we need is that of the exterior derivative dtf> of a 1-form <f>. 

4.4 Definition Let be a 1-form on a surface M. Then the exterior de- 
rivative d<f> of <!> is the 2-form such that for any patch x in M , 

d<j>{x u , x v ) = — (<f>(x v )) — — (0(x„)). 
du dv 

As it stands, this is not yet a valid definition; there is a problem of con- 
sistency. What we have actually defined is a form d x <f> on the image of each 
patch x in M. So what we must prove is that on the region where two patches 
overlap, the forms d x <f> and dy<f> are equal. Only then will we have obtained 
from <f> a single form dfy on M. 

4.5 Lemma Let 4> be a 1-form on M. If x and y are patches in M, then 
d x <t> = d y <t> on the overlap of x{D) and y(E). 

Proof. Because y u and y v are linearly independent at each point, it suf- 
fices by Lemma 4.2 to show that 

(dy<l>) (y u ,y v ) = (dx0)(yu,ye). 

Now as in Corollary 3.4, we write y = x(u,v) and, as in an earlier exercise, 
deduce by the chain rule that 

_ du dv 

du du 

(1) 
_ du dv 

dv dv 

where x„ and x„ are henceforth evaluated on (u,v). Then by Lemma 4.2, 

(40) (y«,y«) = J (d x <i>) (x u ,x v ) (2) 

where J is the Jacobian (du/du)(dv/dv) — (du/dv)(dv/du). Thus it is 
clear from Definition 4.4 that to prove (dy<j>) (y„,y„) = id^) (y«,y„), all we 
need is the equation 

± (*.)-£<*.) -/{£<*.)-£ <*.)}. (3) 

It suffices to operate on (d/du)(<f>y v ), for merely reversing u and v will 
then yield (d/dv)(<f>y u ). Since (3) requires us to subtract these two deriva- 



Sec. 4] DIFFERENTIAL FORMS ON A SURFACE 155 

tives, we can discard any terms that will cancel when u and v are everywhere 
reversed. 

Applying <f> to the second equation in (1) yields 

*(y.) = 0(x„) -^ + tf>(x e ) /. 

Hence by the chain rule, 

where in accordance with the remark above we have discarded two sym- 
metric terms. Next we use the chain rule— and the same remark — to get 

s^>-(s<*-)£+-)s+fe ( ^s + -)s-<« 

Now reverse u and v in (5) (and also u and v) and subtract from (5) 
itself. The result is precisely equation (3). | 

It is difficult to exaggerate the importance of the exterior derivative. We 
have already seen in Chapter I that it generalizes the familiar notion of dif- 
ferential of a function, and that it contains the three fundamental deriva- 
tive operations of classical vector analysis (Exercise 8 of Chapter I, Section 
6). In Chapter II it is essential to the Cartan structural equations (Theorem 
8.3). Perhaps the clearest statement of its meaning will come in Stokes 
theorem (6.5), which could actually be used to define the exterior deriva- 
tive of a 1-form. 

On a surface, the exterior derivative of a wedge product displays the 
same linear and Leibnizian properties (Theorem 6.4 of Chapter I) as in 
E ; see Exercise 3. For practical computations these properties are apt to 
be more efficient than a direct appeal to the definition — compare the dis- 
cussion of the Euclidean case on page 25. Examples of this technique appear 
in subsequent exercises. 

The most striking property of this notion of derivative is that there are 
no second exterior derivatives: Wherever forms appear, the exterior deriva- 
tive applied twice always gives zero. For a surface, we need only prove this 
for O-forms, since even for a 1-form <j>, the second derivative d(d<}>) is a 
3 -form, hence is automatically zero. 

4.6 Theorem If / is a (differentiate) real-valued function on M, then 
d(df) = 0. 

Proof. Let yp = df, so we must show drp = 0. It suffices by Lemma 4.2 
to prove that for any patch x in M we have (df) (x u ,x„) = 0. Now using 
Exercise 4 of Section 3 we get 



156 CALCULUS ON A SURFACE [Chap. IV 



*(0 = df(x u ) = x u [f) = %- (fx) 

du 



and similarly 



dv 
Hence 

#(»., x.) = L (**.) - > (+*„) = £<£> - ^M = o | 

Many computations and proofs reduce to the problem of showing that 
two forms are equal. As we have seen, to do so it is not necessary to check 
that the forms have the same value on all tangent vectors. In particular, 
if x is a coordinate patch, then 

(1) for 1-forms on x(D): </> = yp if and only if <£(x„) = \j/{x u ) and 
<f>(x v ) = i//(x v ); 

(2) for 2-forms on x(D): \i = v if and only if n(x u , x v ) = v(x u , x v ). 
(To prove these criteria we express arbitrary tangent vectors as linear 
combinations of x„ and x„.) More generally, x u and x„ maybe replaced by 
any two vector fields that are linearly independent at each point. 

Let us now check that the rigorous results proved in this section are con- 
sistent with the rules of operation stated in Chapter I, Section 6. 

4.7 Example Differential forms on the plane E 2 . Let U\ = u and u 2 = v 
be the natural coordinate functions, and U\,U 2 the natural frame field on 
E . The differential calculus of forms on E 2 is expressed in terms of Ui and 
Ui as follows : 

If / is a function, <j> a 1-form, and t\ a 2-form, then 

(1) <t> — /i dui + f 2 du 2 , where /»• = 4>(Ut). 

(2) rj = gduidih, where g = r](Ui,U 2 ). 

(3 ) for \[/ = (ft dui + g 2 du2 and <f> as above, 

A ^ = C/i02 — fnQi)dui du 2 . 

(4) df = M. du, + P- du 2 . 

dui du 2 



(5) d<f> = ( — — J dui du 2 (<f> as above). 

\dui du 2 / 

For the proof of these formulas, see Exercise 4. 

Similar definitions and coordinate expressions may be established on any 
Euclidean space. In the case of the real line E 1 , the natural frame field 



Sec. 4] DIFFERENTIAL FORMS ON A SURFACE 157 

(Definition 2.4 of Chapter I) reduces to the single unit vector field Ui for 
which Ui[f] = df/dt. All p-forms with p > 1 are zero, and if <j> is a 1-form, 
then<£ = <f>{Ui) dt. 

Some examples of forms will appear in subsequent exercises; however, 
many more will occur naturally throughout Chapters VI and VII, where 
their properties have direct geometric meaning. 



EXERCISES 

1. If <f> and ^ are 1 -forms on a surface, prove that <j> a rp = — \f/ a <j>. 
Deduce that <f> a <j> = 0. 

2. A form 4> such that dxf> = is said to be closed. A form <j> such that 
<f> = d£ for some form £ is exact. (So if <p is a p-form, £ is necessarily a 
(p — l)-form.) Prove 

(a) Every exact form is closed. 

(b) No 0-form is exact, and on a surface every 2-form is closed. 

(c) Constant functions are closed 0-forms. 

3. Prove the Leibnizian formulas 

d(fg) = dfg+fdg d(j<f>) = df A <t> + f d* 

where/ and g are functions on M, and <f> is a 1-form (Hint: By definition 
(fo)(v P ) = f(p)<t>(y P ); hence f<f> evaluated on x u is /(x)<£(x„).) 

4. (a) Prove formulas (1) and (2) in Example 4.7 using the remark pre- 

ceding Example 4.7. (Hint: Show (duidu2)(Ui, U 2 ) = 1.) 
(b) Derive the remaining formulas using the properties of d and the 
wedge product. 

5. If / is a real-valued function on a surface, and g is a function on the 
real line, show that 

y P \g(f)] = g'(f)y P U)- 

Deduce that 

dig if)) =g'(f)dj. 

6. If /, g, and h are functions on a surface M, and <t> is a 1-form, prove: 

(a) d(fgh) = ghdf + fh dg + fg dh, 

(b) d(4>f) =fd<i>-<t> a df (tf = /0), 

(c) (df a dg)(v,w) = v[/]w[<7] - v[g]w\f]. 

7. Suppose that M is covered by open sets lli, • • • , m^, and on each "U; 
there is defined a function f { such that/* — /_, is constant on the overlap 



158 CALCULUS ON A SURFACE [Chap. IV 

of lit and C U 7 . Show that there is a 1-form <f> on M such that </> = dfc 
on each ll*. Generalize to the case of 1 -forms fa such that fa — <f>j is 
closed. 

8. Let y: E — > M be an arbitrary mapping of an open set of E 2 into a 
surface M. If is a 1-form on M, show that the formula 

r 

d<f>(y u , y v ) = — (0 y.) - ^ (4>y«) 

is still valid even when y is not regular or one-to-one. 

(Hint: In the proof of Lemma 4.5, check that equation (3) is still 
valid in this case.) 

A patch x in M establishes a one-to-one correspondence between an 
open set D of E 2 and an open set x(D) of M. While we have empha- 
sized the function x: D —> x(D), there are some advantages to empha- 
sizing instead the inverse function x _1 : x(D) — > D. 

9. If x: D — ► M is a patch in M, let u and be the coordinate functions 
of x _1 , so x _1 (p) = (#(p), P(p)) for allp in x(Z>). Show that 

(a) u and are differentiable functions on x(D) such that 

u(x(u,v)) = w, v(x(u,v)) = v. 

These functions constitute the coordinate system associated with x. 

(b) du(x u ) = 1 du{x v ) = 
dv(x u ) = dv(x v ) = 1. 

(c) If ^ is a 1-form and 77 is a 2-form, then 

<t> = f du + g dv where /(x) = <f>(x u ), g(x) = <f>(x v ) 
r\ = h dudv where h(x) = 77 (x M , x„). 

(Hint: For (b) use Ex. 4(b) of IV.3.) 

10. Identify (or describe) the associated coordinate system u, v of 

(a) The polar coordinate patch x(u, v) = (u cos v, u sin v) defined 
on D: u > 0, < v < 2tt. 

(b) The identity patch x(u, v) = (u,v) in E 2 . 

(c) The geographical patch x in the sphere. 



5 Mappings of Surfaces 

To define differentiability of a function from a surface to a surface, we follow 
the same general scheme used in Section 3, and require that all its coordi- 
nate expressions be differentiable. 



Sec. 5] 



MAPPINGS OF SURFACES 



159 




5.1 Definition A function F: M — > N from one surface to another is dif- 
ferentiable provided that for each patch x in M and y in N the composite 
function y _1 Fx is Euclidean differentiate (and defined on an open set of 
E 2 ). F is then called a mapping of surfaces. 

Evidently the function y~Vx is defined at all points (u, v) of D such that 
F(x(u, v)) lies in the image of y ( Fig. 4.29). As in Section 3 we deduce 
from Corollary 3.3 that, in applying this definition, it suffices to check 
enough patches to cover both M and N. 

5.2 Example (1) Let 2 be the unit sphere in E (center at 0) but 
with north and south poles removed, and let C be the cylinder based on the 
unit circle in the xy plane. So C is in contact with the sphere along the 
equator. We define a mapping F: 2 — * C as follows: If p is a point of 2, 
draw the line orthogonally out from the z azis through p, and let F(p) be 
the point at which this line first meets C, as in Fig. 4.30. To prove that F 
is a mapping, we use the geographical patch x in 2 (Example 2.2), and for 
C we use the patch y(w, v) = (cos u, sin u, v). Now 

x(u, v) = (cos v cos u, cos v sin u, sin v), 

so from the definition of F, we get 

F(x(u, v)) = (cos u, sin u, sin v). 

But this point of C is y(u, sin v); hence 

F(x(u, v)) = y(w, sin v). 

Applying y~ to both sides of this equation, we 
find 

(y - Fx) (u, v) = (u, sin v) 

so y~ l Fx is certainly differentiable. (Actually x 
does not entirely cover 2, but the missing semicircle 
can be covered by a patch like x.) We conclude 
that F is a mapping. 




FIG. 4.30 



160 



CALCULUS ON A SURFACE 



[Chap. IV 



(2) Stereographic projection of the punctured sphere 2 onto the plane. 
Let 2 be a unit sphere resting on the xy plane at the origin, so the center of 
2 is at (0, 0, 1 ) . Delete the north -pole n = (0, 0, 2 ) from 2. Now imagine that 
there is a light source at the north pole, and for each point p of 2, let 
P(p) be the shadow of p in the xy plane (Fig. 4.31). As usual, we identify 
the xy plane with E 2 by (pi, p 2 , 0) <-> (pi, p%). Thus we have defined a 
function P from 2 onto E 2 . Evidently P has the form 



\ r r / 



where r and R are the distances fromp and P (p), respectively, to the z axis. 
But from the similar triangles in Fig. 4.32, we see that R/2 = r/(2 — p 3 ); 
hence 



P( PllP! , P 3)=(^-,^-) 

\2 — p 3 2 — pzj 



Now if x is any patch in 2, the composite function Px is Euclidean dif- 
ferentiate, so P: 2 — ► E 2 is a mapping. 

Just as for mappings of Euclidean space, each mapping of surfaces has a 
derivative map. 

5.3 Definition Let F: M — > N be a mapping of surfaces. The derivative 
map F* of F assigns to each tangent vector v to M the tangent vector F* (v) 
to N such that: If v is the initial velocity of a curve a in M, then F* (v) is 
the initial velocity of the image curve F(a) in N (Fig. 4.33). 

Furthermore, at each point p, the derivative map F* is a linear trans- 
formation from the tangent plane T P (M) to the tangent plane T F ( P )(N) 
(see Exercise 13 ) . It follows immediately from the definition that F* pre- 
serves velocities of curves: If a = F (a) is the image in N of a curve a in 
M, then F* (a) = a'. As in the Euclidean case, we deduce the convenient 
property that the derivative map of a composition is the composition of 
the derivative maps (Exercise 14). 




FIG. 4.31 




P(p) 



Sec. 5] 



MAPPINGS OF SURFACES 



161 



F*(v) 




FIG. 4.33 




FIG. 4.34 



The derivative map of a mapping F: M — » N may be computed in terms 
of partial velocities as follows. If x: D — * M is a parametrization in M, lefr 
y be the composite mapping F(x): D —* N (which need not be a parametri- 
zation). Obviously F carries the parameter curves of x to the corresponding 
parameter curves of y. Since F* preserves velocities of curves, it follows at 
once that 

F*(x u ) = y u F*(x v ) = y„ 

Since x„ and x„ give a basis for the tangent space of M at each point of 
x(D), these readily computable formulas completely determine F*. 

The discussion of regular mappings in Section 7 of Chapter I translates 
easily to the case of a mapping of surfaces F : M — > N. F is regular provided 
all of its derivative maps F* p : T P (M) — * T F ( P )(N) are one-to-one. Since 
these tangent planes have the same dimension, we know from linear algebra 
that the one-to-one requirement is equivalent to F* being a linear iso- 
morphism. A mapping F: M —> N that has an inverse mapping F -1 :-^ — > M 
is called a diffeomorphism. We may think of a diffeomorphism F as smoothly 
distorting M to produce N. By applying the Euclidean formulation of the 
inverse function theorem to a coordinate expression y _1 ^x for F, we can 
deduce this extension of the inverse function theorem (7.10 of Chapter I). 

5.4 Theorem Let F: M — > iV be a mapping of surfaces, and suppose that 
F* p : T P (M) — > T F ( P ) (N) is a linear isomorphism at some one point p of M . 
Then there exists a neighborhood 11 of p in M such that the restriction of F 
to 11 is a diffeomorphism onto a neighborhood V of F(p) in N (Fig. 4.34). 



T62 



CALCULUS ON A SURFACE 



[Chap. IV 




/a v-garameter curve 01 y^ 

-a M-parameter curve of^T 
FIG. 4.35 

An immediate consequence is that a one-to-one regular mapping F of M 
onto N is a diffeomorphism. For since F is one-to-one and onto, it has a 
unique inverse function P -1 , and F~ l is a (differentiable) mapping, since 
on each neighborhood V, as above, it coincides with the inverse of the 
diffeomorphism 11 — » V. 

5.5 Example Stereographic projection P: S — > E 2 is a diffeomorphism. 

It is clear from Example 5.2 that P is a one-to-one mapping from the 
punctured sphere S to the plane E 2 . Thus we need only show that P* is 
one-to-one at each point. A minor modification of the geographical patch 
in Example 2.2 yields a parametrization 

\(u, v) = (cos v cos u, cos v sin u, 1 + sin v) 

of all of 2 except the south pole, located at the origin 0. 

Now its geometric definition shows that P carries the w-parameter curves 
of x (circles of latitude) to circles in the plane, centered at the origin, and 
that P carries the v-parameter curves (meridians of longitude) to straight 
lines radiating out from the origin, as shown in Fig. 4.35. 

Indeed these two families of image curves are just the parameter curves 
of y = P(x), and from the formula for P in Example 5.2 we get 



y(u,v) = P(x(u,v)) = 



-ft 



cos v cos u 2 cos 



sin v 



1 - 



v sin u\ 
sin v ) 



Since P* (x„) = y„, P* (x„) = y„, the regularity of P* may be proved by 
computing y u and y„, which turn out to be orthogonal and nonzero, hence 
linearly independent. (At the south pole a different proof is required, 
since x is not a parametrization there (see Exercise 15) . ) We conclude that 
P is a diffeomorphism. 

Differential forms have the remarkable property that they can be moved 
from one surface to another by means of an arbitrary mapping, f Let us 
experiment with a 0-form, that is, a real-valued function /. If F : M —> N is 



t This is not the case with vector fields. 



Sec. 5] 



MAPPINGS OF SURFACES 



163 



a mapping of surfaces and/ is a function on M, there is simply no reasonable 
general way to move / over to a function on N. But if instead/ is a function 
on N, the problem is easy; we pull/ back to the composite function f(F) on 
M. The corresponding pull-back for 1 -forms and 2-forms is accomplished as 
follows. 

5.6 Definition Let F: M — > N be a mapping of surfaces. 

(1) If <^> is a 1-form on N, let F*<f> be the 1-form on M such that 

(F**)(v) = <t>(F*v) 

for all tangent vectors v to M. 

(2) If t] is a 2-form on N, let F*r] be the 2-form on M such that 

(F*,)(v,w) = iK*»v,*W) 
for all pairs of tangent vectors v, w on M (Fig. 4.36). 

When we are dealing with a function / in its role as a 0-form, we shall 
sometimes write F*f instead of f(F), in accordance with the notation for 
the pullback of 1 -forms and 2-forms. 

The essential operations on forms are sum, wedge product, and exterior 
derivative; all are preserved by mappings. 

5.7 Theorem Let F: M —> N be a, mapping of surfaces, and let £ and r\ 
be forms on N. Then 

(1) F*(Z + r,) = F*Z + F* V , 

(2) F*($ a „) = F*£ a F*n, 

(3) F*(dt) = d(F*£). 

Proof. In (1), £ and rj are both assumed to be p-forms (degree p — 0, 1, 2) 
and the proof is a routine computation. In (2), we must allow £ and r\ to 
have different degrees. When, say, £ is a function /, the given formula means 
simply F*(fr)) = f(F)F*(ri). In any case, the proof of (2) is also a straight- 
forward computation. But (3) is more interesting. The easier case when £ 
is a function is left as an exercise (Exercise 8 ) , and we address ourselves to 
the difficult case when £ is a 1-form. 



F*(w) 




FIG. 4.36 



164 CALCULUS ON A SURFACE [Chap. IV 

It suffices to show that for every patch x: D — > M 

(d(F*^))(x u ,x v ) = {F*(di-))(x u , x v ). 

Let y = F(x), and recall that F*{x u ) = y M and F*(x v ) = y v . Thus using 
the definitions of d and F*, we get 

d(F*i;)(x u ,x,) = ±{(F*t)(x,)} - 1 {(F*Z)(x u )} 
du OV 

- k ((( *' )} - 1 (i(y - )) - 

Even if y is not a patch, Exercise 8 of Section 4 shows that this last expres- 
sion is still equal to di-(y u , y v ). But 

dtiju, y„) = dit(F*x u , F*x v ) = (F*(d£))(x u , x v ). 

Thus we conclude that d (F*£) and F* (d£) have the same value on x u , x v . | 

The elegant formulas in Theorem 5.7 are the key to the deeper study of 
mappings. In Chapter VI we shall apply them to the connection forms of 
frame fields to get fundamental information about the geometry of mappings 
of surfaces. 



EXERCISES 

1. Let M and N be surfaces in E 3 . If F: E 3 — > E is a mapping such that 
the image F(M) of M is contained in N, then the restriction of F to 
M is a function F \ M : M — > N. Prove that F \ M is a mapping of sur- 
faces. {Hint: Use Theorem 3.2.) 

2. Let S be the sphere of radius r with center at the origin of E . Describe 
the effect of the following mappings F: S — *■ 2 on the meridians and 
parallels of S. 

(a) F(p) = -p. (b) F(pi,p2,pz) = (ps, Pi, Ps). 



f \ Vf \ (Vl + V* Pi ~ P2 \ 

(c) F{p h p 2 , p,) = I ^ ' -v/2 ' ~ P 7 



Let M be a simple surface, that is, one which is the image of a single 
proper patch x : D — > E 3 . If y : D — > ]V is any mapping into a surface N, 
show that the function F: M —> N such that 

F(x(u, v)) = y(w, y) for all (u, v) in D 

is a mapping of surfaces. (Hint: Write F = yx _1 , and use Corollary 3.3.) 



Sec. 5] MAPPINGS OF SURFACES 165 

4. Use Exercise 3 to construct a mapping of the helicoid H (Ex. 7, IV.2) 
onto the torus T (Example 2.6) such that the rulings of H are carried 
into meridians of T. 

5. If 2 is the sphere || p || = r, the function A : 2 — > 2 such that J. (p) = — p 
is called the antipodal mapping of 2. Prove that A is a diffeomorphism 
and that A* (v p ) = ( — v)_ p . 

6. Let x: D — > M be a coordinate patch in a surface M. For any form 
on M, the form x*(0) on D is called a coordinate expression for 0. 
(When^ is aO-form, that is, a function, then x*(0) = ^(x), so this 
terminology is consistent with that of IV.3.) 

If <j> is a 1-form and v a 2-form, prove 
(a) x*(<£) = <f>(x u )du + <j>(x v )dv. (b) x*(v) = p(x M , x v )dudv. 

(c) x*(<ty) = (A (0 x„) - A (0 xj") dw dw. 

(In practice, instead of substituting in the formula (c), it is usually 
easier to apply the exterior derivative to the formula (a).) 

7. (Continuation). Let x be the geographical patch in the sphere 2. 

(a) If <f> is the 1-form on 2 such that <f>(\ p ) = piv 2 — p 2 v u show that 
0(x u ) = r 2 cos 2 y and <f>(x v ) = 0. Then find the coordinate expres- 
sions for <(> and a\f>. 

(b) Prove that the formula v(\ p , w p ) = p$(viw 2 — v 2 Wx) defines a 2- 
form on 2 and find its coordinate expression. 

8. Let F: M —> N be a mapping, and g a function on N. 

(a) Prove that F preserves directional derivatives in this sense: If v 
is a tangent vector to M, then y\g(F)] = (F*\)\g]. 

(b) Deduce that F*(dg) = d(F*(g)). 

9. If x: D — ► M is a parametrization, prove that the restriction of x to a 
sufficiently small neighborhood of a point (wo, v ) in D is a patch in M. 
(Thus a parametrization may be cut into patches.) 

10. If G: P — > M is a regular mapping onto Af, and #: P — ► iV is an arbitrary 
mapping, then the formula F(G(p)) = #(p) is consistent provided 
G(p) = G(q) implies H(p) = #(q) for all p, q in P. Prove that in 
this case F is a well-defined (differentiable) mapping. 

M >N 

G\ /H 
P 

We shall frequently apply this result in the case where G is a para- 
metrization of M. 



166 CALCULUS ON A SURFACE [Chap. IV 

11. Let x: E 2 —*■ T be the parametrization of the torus given in Example 2.6. 

In each case below, show that the formula 

F(x(u, v)) = x(f(u, v), g(u, v)) 

is consistent (Exercise 10), and describe the resulting mapping 
F:T^>T. (For example, give its effect on the meridians and parallels 
of T.) 

(a) / = 3m, g = v. (c) / = v, g = u. 

(b) / = u + x, g = v + 2tt. (d) / = u .+ v, g = u - v. 
Which of these mappings are diffeomorphisms? 

12. Let F: M-* N be a mapping. Let x be a patch in M and let y = F(x). 

(Note that y lies in N, but need not be a patch. ) If 

a(t) = x(a t (t), oa(0) 

is a curve in M, prove that the image a = F(a) in N has velocity 

_/ dai i s da-t , ■, 

oc = -jT y«(ai, as) + -j7 y.voi, a 2 ). 

13. Deduce from Exercise 12: 

(a) The invariance property needed to justify definition (5.3) of F*. 

(b) The fact that derivative maps F*:T P (M) — > T F ( p) (N) are linear 
transformations . 

" w c 

14. Given mappings M > N > P, let GF: M -^ P be the composite 

function. Show that 

(a) GF is a mapping, (b) (GF)* = G*F*, (c) (GF)* = F*G*, 
that is, for any form £ on N, F*(G*£) = (GF)*(£). (Note the reversal 
of factors for (GF)*; forms travel in the opposite direction from points 
and tangent vectors. ) 

15. For stereographic projection P: 2 — » E 2 , show that the derivative map 
at the origin is essentially just an identity mapping. (Hint: Express 
P near in terms of a Monge patch. ) 

16. (a) Prove that the inverse mapping of stereographic projection 

P: 2 — » E 2 is given by the formula 

p- (m , ,) = (^iil , 

where / = w 2 + v. (Show that both PP~ l and P~ l P are identity 
mappings. ) 

(b) Deduce that the entire sphere 2 can be covered by only two 
patches. (The scheme in IV.l requires six.) 



See. 6] INTEGRATION OF FORMS 167 

6 Integration of Forms 

Differential forms have yet another role in the calculus, which the reader 
probably noticed when they first appeared in Chapter I. In, say, a double 
integral fff(u, v) du dv, it is a 2-form on E 2 that appears after the integral 
signs. In a sense, it is only on Euclidean space that forms are actually 
integrated. But we can easily extend this notion of integration to forms on 
an arbitrary surface — by pulling them back to Euclidean space and then 
integrating. 

Consider first the one-dimensional case. By a curve segment (or l-seg- 
ment) in a surface M we shall mean a "curve" a: [a, b] — > M defined on a 
closed interval in the real line E 1 . (Differentiability for a means that it 
can be extended to a genuine curve on a larger open interval as required by 
Definition 4.1 of Chapter I.) 

Now let <f> be a 1-form on M. The pull-back a*4> of </> to the interval 
[a, b] has the expression /(f) dt, where by the remarks following Example 
4.7, 

f(t) = (a**)(tfi(0) =*(*»(tfi(0)) =<*>(«' (0). 

Thus the scheme described above yields the following definition. 

6.1 Definition Let <£ be a 1-form on M, and let a: [a, 6]^Mbea 1 -seg- 
ment (Fig. 4.37). Then the integral of $ over a is 



f $ = / a*4> = f <f>(a(t)) 



dt 



In engineering and physics, the integral f a <f> is called a line integral, and 
it has a wide variety of uses. For example, let us suppose that a vector field 
V on a surface is a force field, so for each point p of M, V(p) is a force 
exerted at p. Returning to our original idea of curve, we further suppose 
that a: [a, b] — » M describes the motion of a mass point — with a(t) its 
position at time t. What is the total amount of work W needed to move a 
from p = a (a) toq = a (b)? The discussion of velocity in Chapter I, 




graph <£(«') 



FIG. 4.37 



168 



CALCULUS ON A SURFACE 



[Chap. IV 



Section 4, shows that for At small, the route of the curve a from a(t) to 
a(t + At) is approximately the straight line segment described by At a(t). 
Now the moving point is opposed only by the component of force tangent 
to a, that is, 



V(a) 



= \\V(a) || cos * 



(Fig. 4.38). Thus the work done against the force during time At is (ap- 
proximately) force -V (a (0). [a (t)/\\ d (t)\\ ] times distance ||a'(0||A*. 
Adding these contributions over the whole time interval [a, b] and taking 
the usual limit, we get 



W 




= -f V(a(t))-a'(t) 

"a 



dt. 



To express this more simply, we introduce 
the dual 1-form <j> such that for each tangent 
vector w at p, <*>(w) = wF(p). Then, by 
Definition 6.1, the total work is just 



a(t + At) 



w 



■-/. 



We emphasize that this notion of line in- 
tegral — like everything we are doing with 
fig. 4.38 forms — applies without change if the surface 

M is replaced by a Euclidean space or, in- 
deed, by any manifold (Section 8). 

When the 1-form <j> is an exterior derivative df, the line integral J a </> has 
an interesting property which generalizes the fundamental theorem of 
calculus. 

6.2 Theorem Let/ be a function on M , and let a: [a, b] — » M be a 1-seg- 
ment in M fromp = a(a) toq = a(b). Then 



Proof. By definition, 



f df = /(q) - /(p). 

•'a 

/ df = f df(a') dt. 

"a "a 



But 



df(a') = a'lfl = | (fa). 



Sec. 6} INTEGRATION OF FORMS 169 

Hence by the fundamental theorem of calculus, 

L df = f.% (fa) dt = /(a(6)) _ /(a(a)) = /(q) " /(p) - ■ 

The integral j a df is thus said to be path-independent. In the language 
used above, if the force field V has dual 1-form df, the work done depends 
not on where the point a (t) moves, but only on where it starts and finishes. 
In particular, if it follows a closed curve, p = a (a) = a(b) = q, it does no 
(total) work at all. 

Mathematically we look at the preceding theorem roughly as follows: 
the "boundary" of the segment a from p to q is q — p, where the purely 
formal minus sign indicates that a goes from p and to q. Then the integral 
of df over a equals the "integral" of / over the boundary q — p; that is, 
/(q) — /(p). This interpretation will be justified by the analogous theorem 
(6.5), in dimension 2. 

Now a two-dimensional interval is just a closed rectangle R : a ^ u ^ b, 
c ^ v ^ d in E 2 . And a 2-segment in M is a differentiate mapping x: R —> M 
of a closed rectangle into M (Fig. 4.39 ) . (As before, differentiability means 
one can extend x differentiably to an open set containing R. ) 

Although we use the patch notation x, we do not assume that x is either 
regular or one-to-one. The partial velocities x u and x„ are still available, 
however, even when x is not a patch. 

If 77 is a 2-form on M, then the pullback x*tz of 77 has, using Example 4.7, 
the coordinate expression h du dv, where 

h = (x*?/)(£/i, C/ 2 ) = rj(x*C/i, x*C/ 2 ) = r?(x M , x„). 

Thus by strict analogy with Definition 6.1 we establish 

6.3 Definition Let tj be a 2-form on M , and let x: R —> M be a 2-segment. 
Then the integral of rj over x is 

JJ v = JJ x * v = J J v ^ Xu ' Xv ^ du dv ' 
The physical applications of this notion of integral are perhaps richer 




FIG. 4.39 



170 



CALCULUS ON A SURFACE 



[Chap. IV 



than those of Definition 6.1, but we must proceed without delay to the 
two-dimensional analogue of Theorem 6.2. 

6.4 Definition Let x: R — > M be a 2-segment in M with R the closed rec- 
tangle a ^ u ^ b, c ^ v ^ d (Fig. 4.40). The edge curves (or edges) of x 
are the 1 -segments a, (3, y, 8 such that 

a(u) = x(u, c) 



0(v) = x(b,v) 
y(u) = x(u, d) 
8(v) = x(a, v) 



d 



(y) 



■ (*) R (/?) 



IS 



a b " 

FIG. 4.40 

Then the boundary dx of the 2-segment x is the formal expression 

dx = a + P — y — 8. 

These four curve segments are what we get by considering the function 
x:R-+M only on the four line segments that comprise the boundary of the 
rectangle R. The formal minus signs before y and 8 in dx remind us that y 
and 8 must be "reversed" to give a consistent trip around the rim of R, and 
thus of x (Fig. 4.41). Then if <f> is a 1-form on M, the integral of </> over the 
boundary of x is defined to be 



/0=/*^+/^-/'0-/ , 0. 

J dx Ja J B J", ->& 



J dx J a •'/S 

The two-dimensional analogue of Theorem 6.2 is then 

6.5 Theorem (Stokes' Theorem) If <f> is a 1-form on M, and x: R — > M 
is a 2-segment, then 



II d<f> = / 



Proof. We shall work on the double integral and show that it turns out 



R 




FIG. 4.41 



Sec. 6] 



INTEGRATION OF FORMS 



171 



to be the integral of <f> over the boundary of x. Combining Definitions 6.3 
and 4.4 we have 

J J d<f> = J J (d4>) (x« , x„) du dv = J J (— (<f>x v ) - — (<t>x u ) J du dv 

Let/ = <t>(x u ) and g = <f>(x v ); this equation then becomes 

[f d4= (f d ± du dv - [f d 4du dv (1) 

JJ X JJr du JJr dv 

Now we treat these double integrals as iterated integrals. Suppose the 
rectangle R is given by the inequalities a^u^b, c^v^d. Then inte- 
grating first with respect to u, we find 

[[ ^ dudv = [ I(v)dv, where I(v) = [ ^ (u, v) du. 

JJr dU J c J a dU 

In the partial integral defining I (v), v is constant, so the integrand is just 
the ordinary derivative with^ respect to u. Thus the fundamental theorem 
of calculus applies to give 

I(v) = flf(M) - 9(a,v). 

(Fig. 4.42). Hence 

[[ d 4 dudv = [ g(b, v) dv - f g(a, v) dv. (2) 

JJr dU J c J c 

Again we work on the first integral. Now by definition 

g(b,v) = 0(x„(6, v)). 

But x v (b, v) is precisely the velocity /3'(y) of the "right side" curve in 
dx. Hence by Definition 6.1, 

J g(b,v)dv = f 4>(p'(v))(kr == / <f>. 



(a,v) 



(b, v) 



FIG. 4.42 



172 CALCULUS ON A SURFACE [Chap. IV 

A similar argument shows that the second integral in (2) is f& <f>; hence 

ff m £* , *-f,*-f,+ (3) 

In the same way — but integrating first with respect to v\ — we find 

//.s**-/ T *-/.* (4) 

Assembling the information in (1), (3), and (4) we obtain the required 
result: 



//.*-{i* -/*}"{/,♦ -/.♦}"/.♦ 



i 



Stokes' theorem may be considered a two-dimensional formulation of 
the fundamental theorem of calculus; it ranks as one of the most useful 
results in mathematics. Alternative formulations of the theorem and ex- 
tensive applications may be found in texts on advanced calculus or applied 
mathematics; we shall use it to study the geometry of surfaces. 

The line integral /« is not particularly sensitive to reparametrization of 
a; all that matters is the direction in which the route of a is traversed. 
The following lemma uses the notation of Exercise 10, II.2. 

6.6 Lemma Let a (h) : [a, b] — > M be a reparametrization of a curve seg- 
ment a: [c, d] — ► M . For any 1-form <f> on M 



/ ♦- 

Ja(h) 



I <j) if h is orientation-preserving 

•'a 

I 4> if h is orientation-reversing 

J a 

Proof. Since a(h) has velocity 

a(h)' = j-a'(h), 

we have 

dh 



f <t> = f 4>(a{hY) du= f 0(«'(A)) 



1 du 
du 



Now we use the theorem on change of variables in an integral. If h is 
orientation -preserving, then h(a) = c and h(b) = d, so the integral above 
becomes 



/ 4>(a) du = I <j>. 



Sec. 6] INTEGRATION OF FORMS 173 

But in the orientation-reversing case, h(a) = d and h(b) = c, which gives 

] <*>(«') du = -] <f>(a) du = -J 4> I 

This lemma lets us give a concrete interpretation to the formal minus 
signs in the boundary dx = a + & — 7 — 5 of a 2-segment x. For any 
curve £: [t , U] — > M, let — £ be any orientation-revemngf reparametrization 
of £, say 



Thus by the lemma, 



(-0(0 = €(*> + <i - 0. 






and if x is a 2-segment, then 

[<l>=l<t>+[<1>-f<l>-l<t>=l<f>+f<t>+[ <{>+ [ <f>. 

J dx J a J0 Jy Ji J a J P J-y J-S 



EXERCISES 

1. If a = (ai, a 2 ) is a curve in E 2 and is a 1-form, prove this computa- 
tional rule for finding <f>(a) dt: Substitute u = ai(t) and v = a 2 (t) 
in a coordinate expression <f> = / (u, v) du + g (u, v) dv. 

2. Consider the curve segment a: [— 1, 1] — > E 2 such that a(t) = (t,t 2 ). 

(a) If <f> = v du + 2wi; dv, compute /« 0. 

(b) Find a function / such that df = <j> and check Theorem 6.2 in this 
case. 

3. Let be a 1-form on a surface M. 

(a) If <f> is closed, show that j dx </> = for every 2-segment in M . 

(b) If is ezad, show (more generally) that 



s/.*-° 



for any "cycle" of curve segments «i, • • • , 
a k (<xk+i = cti ) such that a; ends at the start- 
ing point of cti+i (Fig. 4.43). (Closed means 
d4> = 0, ezaef means = d/ ; see Ex. 2 of 

IV.4.) 

4. The 1-form 




FIG. 4.43 



174 CALCULUS ON A SURFACE [Chap. IV 

_ u dv — v du 

* ~ U 2 + V 2 

is well-defined on the plane E 2 with the origin (0, 0) removed. Show: 

(a) yp is closed, but not exact. {Hint: Integrate around the unit circle 
and use Exercise 3.) 

(b) if \f/ is restricted to, say, the right half -plane, u > 0, then \]/ be- 
comes exact. 

5. (Continuation) . It follows from Ex. 12 of II. 1 that every curve a in 
E that does not pass through the origin can be written in the polar 
form 

a(t) = (r(t) cos #(t),r(t) sintK*)). 

Prove that for every closed curve a, (l/2ir) J a ^ is an integer. This 
integer is called the winding number of a; it represents the total alge- 
braic number of times a has gone around the origin in the counter- 
clockwise direction. 

6. Let x be a patch in a surface M. For a curve segment 

a(t) = x(ai(0, O2(0), a ^ t ^ b, 
show that 

/.♦-/>*•>£+♦<*>£)* 

where x u and x v are evaluated on (a iy a 2 ). (This generalizes Ex. 1, since 
we can use the identity patch x(w, v) = (u, v) on E .) 

7. Let a be the closed curve 

a(t) = x(mt, nt), S t S 2x 

in the torus T (see Ex. 11 of IV.3.) Compute 

(a) Ja £, where £ is the 1-form on T such that £(x„) = 1, and £(x c ) = 0. 

(b) /« r?, where ?7 is the 1-form such that rj (x u ) = and t}(x v ) = 1. 
If 7 is an arbitrary closed curve, / 7 £ /2ir gives the total number of 

times 7 travels around the torus in the general direction of the parallels, 
while j y r)/2ir gives a similar measurement for the direction of the 
meridians. This suggests the commonly used notation £ = d&, t\ = d<p, 
where t? and <p are the (multivalued ! ) longitude and latitude functions 
on T; however, see Exercise 13. 

8. Let F: M — ► N be a mapping. Prove: 

(a) If a is a curve segment in M and ^ is a 1-form on N, then 



J a J F(a) 



Sec. 6] INTEGRATION OF FORMS 175 

(b) If x is a 2-segment in M and v is a 2-form on N, then 

f F*p = [ v. 

Jj. J F(x) 

9. Let x: R -> 2 be the 2-segment in the sphere 2 obtained by restricting 
the geographical patch 

x(w, v) = (r cos v cos u, r cos t; sin u, r sin v) 

to the rectangle R: ^ u, v ^ x/2. Find explicit formulas for the edge 
curves a, j8, 7, 5 of x, and show these curves and the image x(R) on a 
sketch of 2. 

10. Let x: R -» If be a 2-segment denned on the rectangle 

#:0 ^ w,v ^ 1. 

If <£ is the 1-form on M such that 

<t>(x u ) = u + v and <£(x„) = wv, 

compute jj x d4> and J 3x </> and check the results by Stokes' theorem. 
{Hint: x*(cty) = d(x*0) = (» - l)du d».) 

11. Same as Exercise 10, except that R: ^ u ^ tt/2, ^ v ^ x, and 
0(x u ) = u cos v, <£(x,,) = v sin w. 

12. A closed curve a in M is homotopic to a constant provided there is a 
2-segment x: R -> M for which (a) a is, in fact, the a edge curve of x, 
(b) = 5, and (c) 7 is a constant curve (Fig. 4.44). (Suppose 
R:a ^ u ^ b, c ^ v ^ d. Then as v varies from c to d, the closed 
w-parameter curve, v = vo, of x varies smoothly from a to the constant 
curve 7.) Prove that every closed curve in E 2 is homotopic to constant. 




^ 



FIG. 4.44 FIG. 4.45 

13. Let <j> be a closed 1-form and a a closed curve. Prove that J a <£ = 
if either 
(a) <f> is exact, or (b) a is homotopic to a constant. 

Deduce that on the torus T meridians and parallels are not homotopic 
to constants, and the closed forms £ and 77 (Exercise 7) are not exact. 



176 CALCULUS ON A SURFACE [Chap. IV 

A surface M for which every closed curve is homotopic to a constant 
is said to be simply connected. Thus the plane is simply connected 
(Ex. 12 ), but a torus, or a plane with even a single point removed, is not 
(Ex. 7 and 5). Roughly speaking, a simply connected surface has no 
holes in it, and any four curves a, ft 7, 8 linked together as in Fig. 4.45 
are in fact the boundary curves of some 2-segment.f Use this assertion 
to prove: 
14. On a simply connected surface, the integral of a closed 1-form <f> is path- 
independent. (That is, /„ 4> is the same for all a. with the same end 
points. ) 

15. On a simply connected surface, every closed 
1-form is exact. (Hint: Fix a point p in M and 
define /(p) = j s <f> for any curve segment 
frompo to p (Fig. 4.46). To show df(y) = 
4>(y) for a tangent vector v at p, prove that 
if a is a curve with initial velocity v, then 



P = a(0) 




/(«(*)) =/(p) + f <t>{*\u))du) 



FIG. 4.46 



7 Topological Properties of Surfaces 

We now discuss some of the very basic properties that a surface may possess. 

7.1 Definition A surface M is connected provided that for any two points 
p and q of M there is a curve segment in M from p to q. 

Thus a connected surface M is all in one piece, since one can travel from 
any point in M to any other without leaving M . Most of the surfaces we 
have discussed so far have been connected; the surface M:z 2 — x 2 — y 2 = 1 
(hyperboloid of two sheets) is not connected. Connectedness is a mild and 
reasonable condition and might well be included in the definition of surface. 

7.2 Definition A surface M is compact provided that M can be covered 
by the images of a finite number of 2-segments in M . 

Roughly speaking, compactness means that the surface is finite in size. 
For example, spheres are compact, since if we use the formula for x(u, v) 
in Example 2.2 on the closed rectangle 

R: -ir ^ u ^ 7T, -tt/2 S v S x/2, 

t For a systematic account of simple connectedness, see pp. 157-165 of Lef schetz 
[8], where it is shown that spheres are simply connected. 



Sec. 7] TOPOLOGICAL PROPERTIES OF SURFACES 177 

then 2 is the image of this single 2-segment. Similarly the torus of revolu- 
tion (Example 2.6), or any closed surface of revolution, is compact. 

The proof of the following lemma uses this fundamental fact : If / is a 
continuous real-valued function defined on a closed rectangle 

R: a rg u ^ b, c ^ v ^ d, 

then / takes on its maximum at some point of R. 

7.3 Lemma If / is a continuous function on a compact surface M, then 
/ takes on maximum at some point of M. (Obviously we can also replace 
maximum by minimum.) 

Proof. By definition there exist a finite number of 2-segments 

Xi.Ri^M (1 ^ i S k) 

whose images cover all of M . Since each x, is differentiate, it is also con- 
tinuous, so each composite function /x» : R % ■— > R is continuous. Thus by the 
remark above, for each index i, there is a point (m», «,-) in Ri where the 
function /x t takes its maximum. Let, say, /(xi(tti, Vi)) be the largest of 
this finite number A: of maximum values. We assert that / takes on its 
maximum value at the point m = xi(wi, Vi). In fact, we shall prove that if 
p is any point of M, then /(m) ^ /(p). Since the 2-segments x x , • • • , x k 
cover M , there is an index i such that p = Xj(w, v). But then by the pre- 
ceding construction, 

/(m) = /(xx(M!, v^) ^ f(xi(Ui, Vi)) ^ f(xi(u, v)) = /(p). | 

This very useful result can be applied to prove noncompactness. For 
example, no cylinder C (as in Example 1.5) is compact, since the coor- 
dinate function z on C gives the height z(p) of each point p above the xy 
plane, and thus has no maximum value on C. 

However, Definition 7.2 is a little trickier than it looks. Consider, for 
example, the open unit disc D : x 2 + y 2 < 1 in the xy plane. Now, D is a 
surface and has finite area x. But D is not compact : It suffices to note that 
the continuous function / = (1 — x 2 — y 2 )~ l does not have a maximum on 
D. In general, a compact surface cannot have any open edges, as D does. 
It must be smoothly closed up everywhere — as well as finite in size — like 
a sphere or torus. 

Roughly speaking, an orientable surface is one that is not twisted. Of 
the many equivalent formulations of orientability the one that follows is 
perhaps the simplest. 

7.4 Definition A surface M is orientable provided there exists a 2-form y. 
on M which is nonzero at each point of M. 

(A 2-form is zero at a point p if it is zero on every pair of tangent vectors 



178 CALCULUS ON A SURFACE [Chap. IV 

at p. ) Thus the plane E 2 is orientable, since du dv is a nonvanishing 2-form. 
Although simple, this definition is somewhat mysterious, so we shall prove 
a more intuitive criterion, 

7.5 Theorem A surface M in E 3 is orientable if and only if there exists a 
normal vector field Z on M that is nonzero at each point of M. 

Proof. We use the cross product of E 3 to convert normal vector fields into 
2-forms, and vice versa. For Z as above, define a 2-form n on M as follows: 
For any pair v, w of tangent vectors to M at p, let 

m(v, w) = Z(p)»v X w. 

Standard properties of the cross product show that \i is, in fact, a nonvanish- 
ing 2-form on M. Thus M is orientable. 

Conversely, suppose that M is orientable, with n a nonvanishing 2-form. 
If v, w is a linearly independent pair of tangent vectors at p, then 

M (v, w) *0, 

for otherwise, n would be zero at p. Now define 

v X w 



Z(p) = 



/*(v, w) * 



This formula has the remarkable property that it is independent of the 
choice of v, w at p. Explicitly, for any other such pair v, \v, it follows from 
Lemma 4.2 and the analogous formula for cross products that 

v X w _ v X w 

/«(v, w) m (v, w) ' 

We have thus obtained a well-defined Euclidean vector field on all of M . 
Again the properties of the cross product show that Z is everywhere normal 
to M, but never zero. | 

Thus it follows from Lemma 3.8 that every surface in E 3 that can be 
defined implicitly is orientable. For example, all cylinders, surfaces of 
revolution, and spheres (in fact, all quadric surfaces) are orientable. How- 
ever, nonorientable surfaces do exist in E 3 . The simplest example is the 
famous Mobius band M , which can be made from a strip of paper by giving 
it a half twist, then gluing its ends together. (The formal construction of a 
particular Mobius band is given in Exercise 7.) M is nonorientable, since 
every normal vector field Z on M must somewhere be zero. To see this, 
let 7 be a closed curve as indicated in Fig. 4.47, with 7(0) =7(1) = p. If 
we assume that Z is never zero, then the twist in M forces the contradiction 
Z(7(l)) = —Z(y(0)), since the function t — > Z(y(t)) is differentiate 
(that is, Z varies smoothly as it moves around 7). 



Sec. 7] 



TOPOLOGICAL PROPERTIES OF SURFACES 



17? 



The three properties discussed in this section — connectedness, compact- 
ness, and orientability — are topological properties: It is possible to define 
them using only open sets and continuous functions, with no differentiabil- 
ity considerations at all. Using these more general definitions, a more con- 
ceptual proof can be given of the following result. 

7.6 Theorem Let M and N be surfaces in E such that M is contained in 
N. If M is compact, and N is connected, then M = N. 

(If N were not connected, it might consist of two surfaces, of which M 
is one. Similarly the result fails if M is not compact; consider the case of 
an open disc M in the xy plane N. ) 

Proof. Exercise 15 of Section 3 shows that M is an open set of N. We 
assume that M does not fill all of N, and deduce a contradiction. By assump- 
tion there is a point n of N that is not in M. Let m be a point of M . Since 
N is connected, there is a curve segment a in N from 



a(0) =m 



to 



:(D =n. 



Let t be the least upper bound of those numbers t such that a(t) is in M- 
We assert that the point p* = a(t*) is in M (Fig. 4.48). 

To prove this, consider the real-valued function f on M such that at each 



Z(y(0)) 




Z(y(D) 




FIG. 4.48 



1 80 CALCULUS ON A SURFACE [Chap. IV 

point p of M, /(p) is the Euclidean distance d(p*, p) from p* to p. Now 
/ ^ is continuous, in the sense that for each patch x in M the composite 
function /(x) is continuous. Since M is compact, Lemma 7.3 applies to 
show that / takes a minimum at some point of M. By definition of least 
upper bound, there are numbers t < t*, arbitrarily close to t*, such that 
a(t) is in M. Since a is continuous (being differentiable) the corresponding 
distances d(p*, a(t)) become arbitrarily small; hence the minimum value 
of / can only be zero. Thus the only possible point at which / can be a 
minimum is p* itself, which means that p* is in the domain M of /. 

Since M is an open set of N, it follows that every point of N near enough 
to p* is also in M. Thus if t > t* is near enough to t*, a(t) must still be 
in M — a contradiction to the definition of t*. | 

EXERCISES 

1. Decide which of the following surfaces are compact and which are 
connected : 

(a) A sphere with one point removed. (c) The surface in Fig. 4.10. 

(b) The region z > in M: z = xy. (d) M : x 2 + y + z 6 = 1. 
(e) A torus with the curve a (t) = x(t, t) removed. (See Ex. 2 of IV.3.) 

2. Let Fbea mapping of a surface M onto a surface N. Prove: 

(a) If M is connected, then N is connected. 

(b) If M is compact, then N is compact. 

3. Let F: M — > N be a regular mapping. Prove that if N is orientable> 
then M is orientable. 

4. Let / be a differentiable real-valued function on a connected surface M. 
Prove that: 

(a) If df = 0, then / is constant. 

(b) If/ is never zero, then either/ > or/ < 0. 

5. (a) Prove that a connected orientable surface has exactly two unit 

normal vector fields, which are negatives of each other. We denote 
these by ±U. (Hint: Use Ex. 4.) 
(b) If M is a nonorientable surface, prove that any point of M is 
contained in a connected orientable region. (Thus, even on a 
nonorientable surface, unit normal vector fields exist locally.) 

6. Let F: M — > N be a regular mapping. Prove this generalization of 
Theorem 7.6: If M is compact and N is connected, then F carries M 
onto N. 

7. A Mobius band M (Fig. 4.47) can be constructed as a ruled surface 

x(u, v) = j8(m) + v&(u), -% ^ v ^ \, 



Sec. 8] MANIFOLDS 181 

where 

/3(w) = (cos u, sin u, 0) 

a(«) = (cos|)/3( W ) + (sin|)t/ 3 . 

(The ruling L makes only a half -turn as it traverses the circle once. ) 

(a) Compute 

E = V - + [1 + v cos (u/2)] 8 , F = 0, G = 1, 
4 

and deduce as in Exercise 2 of Section 2 that x is regular. 

(b) Show the w-parameter curve, v = £, on a sketch of M. Prove that 
the w-parameter curves are closed and (0 excepted) have period 
4tt. 

8. Let M * be the surface obtained by removing the central circle j8 from 
the Mobius band in the last exercise. Is M* connected? Orientable? 

9. (Counterexamples). Give examples to show that the following are all 
false : 

(a) Converses of (a) and (b) of Exercise 2. 

(b) Exercise 3 with F not regular. 

(c) Converse of Exercise 3. 

10. A surface M in E 3 is closed in E 3 provided the points of E 3 not in M 
constitute an open set of E 3 . (If p is not in M, there is an e-neighborhood 
of p that does not meet M. ) Show that : 

(a) Every surface in E 3 that can be described in the implicit form 
M : g = c is closed in E . 

(b) Every compact surface in E is closed in E . 

11. (Boundedness). A surface M in E 3 is bounded provided there is a num- 
ber R such that || p || ^ R for all points p of M. (Thus M lies inside a 
sphere.) Prove that a compact surface in E is bounded. 

The last two exercises have shown that a compact surface in E 3 is closed 
and bounded; the converse follows from a fundamental topological theorem. 

12. Prove Theorem 7.6 assuming M is merely closed in E 3 (instead of 
compact). 

13. In each case, decide whether the surface M: g = 1 is compact, or 
connected : 

(a) g = x 2 - y s + z\ (c) g = £ + xy\ 

(b) g = x 4 - y + z. (d) g= (x 2 + y 2 - 4) 2 + (z - 4) 2 . 



1 82 CALCULUS ON A SURFACE [Chap. IV 

14. Prove that every surface of revolution M is connected and orientable, 
but that M is compact if and only if its profile curve C is closed. 



8 Manifolds 

Surfaces in E are a matter of everyday experience, so it is reasonable to 
try to investigate them mathematically. But examining this concept with 
a critical eye, we may well ask if there could not be surfaces in E 4 - • • or 
E n • • • or even surfaces that are not in any Euclidean space at all. To devise 
a definition for such a surface, we must rely not on our direct experience of 
the real world, but on our mathematical experience of surfaces in E 3 . Thus 
we shall strip away from the basic definition (1.2) every feature that 
involves E in any way. What is left will be just a surface. 

To begin with, a surface will be a set M: a collection of any objects 
whatsoever, not necessarily points of E 3 . An abstract patch in M will now be 
just a one-to-one function x: D — > M from an open set D of E 2 into the set 
M. There is, as yet, no way to say what it means for such a function to be 
differentiable. But all we need to get a workable definition of surface is the 
smooth overlap condition (Corollary 3.3). To prove this now is a logical 
impossibility, so in the usual fashion of mathematics, we make it an axiom. 

8.1 Definition A surface is a set M furnished with a certain collection (P 
of abstract patches in M such that: 

(1 ) The images of the patches in the collection (P cover M. 

(2 ) For any two patches x, y in the collection (P, the composite functions 
y _1 x and x _1 y are Euclidean differentiable (and defined on open sets of E 2 ). 

This definition generalizes Definition 1.2: A surface-in-E 3 is a surface. 
But there are vast numbers of surfaces that can never be found in E 3 . 

8.2 Example The projective plane 2. Starting from the unit sphere 2 
in E , we construct the projective plane 2 by identifying antipodal points of 
2; that is, by considering p and — p to be the same point (Fig. 4.49). For- 



FIG. 4.49 



Sec. 8] MANIFOLDS 183 

mally, this means the set 2 consists of all antipodal pairs {p, — p} of points 
of the sphere. (Order is not important here; that is, {p, — p} = { — p, p}.) 

To make 2 a surface — and later to study it — we use two functions: the 
antipodal mapping A(p) = — p of the sphere 2, and the projection P(p) = 
{p, -p} of 2 onto 2. Note that PA = P, and (J): P(p) = P(q) if and 
only if either q = p or q = —p. 

Let us call a patch x in 2 "small" if the Euclidean distance between any 
two of its points is less than 1. If x: D — ► 2 is a small patch, then the com- 
posite function P(x) : D — > 2 is one-to-one, and is thus an abstract patch 
in 2. The collection of all such abstract patches makes 2 a surf ace — the first 
condition in Definition 8.1 is clear, and we merely outline the proof of the 
second. 

Suppose that P(x) and P(y) overlap in 2; that is, their images have a 
point in common. If x and y overlap in 2, show that (Py) _1 (Px) = y -1 x, 
which by Corollary 3.3 is differentiable and defined on an open set. (Hint: 
Use smallness and %.) On the other hand, if x and y do not overlap, replace 
y by A(y). Then x and ^4.(y) do overlap, so the previous argument applies. 

Conclusion: The projective plane 2 is a surface. f 

To emphasize the distinction between a surface in E 3 and the general 
concept of surface defined above, we shall sometimes call the latter an 
abstract surface. Note that E 2 is an abstract surface if it is furnished with the 
single patch x(u, v) = (u, v). 

To get as many patches as possible in an abstract surface M, it is custom- 
ary to enlarge the given patch collection (P to include all abstract patches 
in M that overlap smoothly with those in (P. In working with M, these 
are the only patches we can use. We emphasize that abstract surfaces Mi 
and M 2 with the same set of points are nevertheless different surfaces if 
their (enlarged) patch collections (Pi and (P 2 are different. 

There is essentially only one problem to solve in establishing the calculus 
of an abstract surface M, and that is to define the velocity of a curve in M. 
For everything else — differentiable functions, curves themselves, tangent 
vectors, tangent vector fields, differential forms, and so on — the definitions 
and theorems given for surfaces in E 3 apply without change. (It is neces- 
sary to tinker with a few proofs to take care of the new Definition 8.3, but 
no serious problems arise.) The velocity of a curve fails us in the abstract 
case, since before it consisted of tangent vectors to E 3 , and now E 3 is gone. 

It makes not the slightest difference what we define the velocity a (t) 
to be — provided the new definition leads to the same essential properties 
as before. The directional derivative property (Lemma 4.6 of Chapter I) 
is what is needed. 

t The terminology in this example derives from projective geometry, however, 
Exercise 2 shows that 2 can more aptly be described as a twisted sphere. 



134 CALCULUS ON A SURFACE [Chap. IV 

8.3 Definition Let a: I — > M be a curve in an abstract surface M. For 
each t in /, the velocity vector a (t) is the function such that 

«'(0W = ^ (0 

for every differentiable real-valued function / on M. 

Thus a (t) is a real-valued function whose domain is the set of all differ- 
entiable functions on M. This is all we need to generalize the calculus for 
surfaces in E to the case of an abstract surface. 

The reader may feel he has gone far enough in the direction of abstrac- 
tion, but in one more step we shall have gone all the way. 

We now have a calculus for E" (Chapter I) and a calculus for surfaces. 
These are strictly analogous, but analogies in mathematics (although use- 
ful at first) are, in the long run, annoying. What we need is a single calculus, 
of which these two will be special cases. The most general object on which 
calculus can be conducted is called a manifold. It is simply an abstract 
surface of arbitrary dimension n. 

8.4 Definition An n-dimensional manifold M is a set furnished with a 
collection (P of abstract patches (one-to-one functions x: D — ► M, D an 
open set in E n ) such that 

(1 ) M is covered by the images of the patches in the collection (P. 

(2) For any two patches x, y in the collection (P, the composite func- 
tions y _1 x and x~*y are Euclidean-differentiable (and defined on open sets 
inE n ). 

Thus a surface (Definition 8.1) is the same thing as a two-dimensional 
manifold. The Euclidean space E" is a very special n-dimensional manifold; 
its patch collection consists only of the identity function. 

To keep this definition as close as possible to that of a surface in E 3 , we 
have deviated slightly from the standard definition of manifold in which it 
is usually the inverse functions x _1 : x (D ) — » D that are axiomatized. 

The calculus of an arbitrary n-dimensional manifold M is defined in the 
same way as in the special case, n = 2, of an abstract surface. Differentiable 
functions, tangent vectors, vector fields, and mappings are gotten exactly 
as before: We need only replace i = 1, 2 by i = 1, 2, • • • , n. Differential 
forms on a manifold M have the same general properties as in the case 
n = 2, which we have explored in Sections 4, 5, and 6. But there are p- 
forms for ^ p ^ n, so when the dimension n of M is large, the situation 
becomes rather more complicated than for n = 2, and more sophisticated 
techniques are called for. 

Whenever calculus appears in mathematics and its applications, mani- 
folds will also be found, and higher -dimensional manifolds turn out to be 



Sec. 8] 



MANIFOLDS 



185 




FIG. 4.50 

important in problems (both pure and applied) that initially seem to in- 
volve only dimensions 2 or 3. For example, we now describe a four-dimen- 
sional manifold that has already appeared, implicitly at least, in this 
chapter. 

8.5 Example The tangent bundle of a surface. If M is a surface, let T (M ) 
be the set of all tangent vectors to M at all points of M . (For the sake of 
concreteness we shall think of M as a surface in E 3 , but M could just as well 
be an abstract surface, or indeed a manifold of any dimension.) Now M 
itself has dimension 2 and each tangent plane T P (M) has dimension 2, so 
T(M) will turn out to have dimension 4. To get the patch collection (P 
that will make the set T(M) a manifold, we shall derive from each patch 
xinla patch x in T(M). Given x: D — > M, let 6 be the open set in E 4 
consisting of all point (p h p 2 , p 3 , p 4 ) for which (p u p 2 ) is in D. Then let 
x: D — * T(M) be the function such that 

Xfal, P2, PZ, Pi) = P»X«(pi, Pi) + P4X B (P1, Pi). 

(In Fig. 4.50 we identify E 2 with the x x Xi plane of E 4 and deal as best we 
can with dimension 4.) 

Using Exercise 3 of Section 3 and the proof of Lemma 3.6, it is not diffi- 
cult to check that (1 ) each such function x is one-to-one, hence is a patch 
in T{M), in the sense of Definition 8.4, and (2) the collection & of all 
such patches satisfies the two conditions in Definition 8.4. Thus T(M) is 
a four-dimensional manifold called the tangent bundle of M. 



EXERCISES 

Show that a surface M is nonorientable if there is a closed curve 
a: [0, 1] — * M and a vector field Y on a such that 

(a) Y and a are linearly independent at each point. 

(b) F(l) = -F(0). 



186 CALCULUS ON A SURFACE [Chap. IV 

2. Establish the following properties of the projective plane 2: 

(a) If P: 2 — ■» 2 is the projection, then each tangent vector to 2 
is the image under P* of exactly two tangent vectors to 2 — of 
the form v p and ( — v)_„. 

(b) 2 is compact, connected, and nonorientable. 

(Hint: For (a), use Ex. 5 of IV.5.) Although the proof is difficult, 
every compact surface in E is orientable — thus 2 is not diffeomorphic to 
any surface in E 3 . 

3. Prove that the tangent bundle (8.5) is a manifold. (If x'and y are 
overlapping patches in M, find an explicit formula for y -1 x.) 

4. If M is the image of a single patch x: E — > M, show that the tangent 
bundle T(M) is diffeomorphic to E 4 . 

5. (Plane with two origins). Let M consist of all ordered pairs of real 
numbers (u, v) and one additional point 0*. Let x and y be the func- 
tions from E 2 to M such that 

x(u, v) = y(u, v) = (u, v) if (u, v) ^ (0, 0), 

but 

x (0, 0) = = (0, 0) and y(0, 0) = 0*. 

Prove that: 

(a) The abstract patches x and y make M a surface. 

(b) M is connected. 

(c) The function F: M — » M is a mapping, where F(0) = 0* and 
F(0*) = 0, but F(p) = p for all other points of M. 

Surfaces such as this one are troublesome to deal with; we eliminate 
them by adding an additional hypothesis to Definition 8.1: For any 
points p 9^ q of M there exist abstract patches x and y in (P such that 
p is in x(D), q is in y (E), and x(D) and y (E) do not meet (Hausdorff 
axiom). 

6. Let V be a vector field on a surface M. A curve a in M is an integral 
curve of V provided a (t) = V(a(t)) for all t. Thus an integral curve 
has at each point the velocity prescribed by V. If a(0) = p, we say 
that a starts at p. 

(a) In the special case M = E 2 , show that the curve t — > (ai(t), a^(t)) 
is an integral curve of v = /i C/i + U Ui starting at (a, b) if 
and only if 

-tj = /i(«i» ai) fai(0) = a 

and 



-77 = Mai, a 2 ) la 2 (0) = b 

at 



Sec. 9] SUMMARY 187 

The theory of differential equations predicts a unique solution for 

such systems. 

(b) Find the integral curve of 

v = —u 2 Ui + uvUz on E 2 

which starts at the point (1, —1). {Hint: The differential equa- 
tions in this case can be solved by elementary methods. Use the 
arbitrary constants in the solution to make the starting point 
(1, -DO 

7. Prove that every vector field V on a surface M has an integral 
curve starting at any given point. (Hint: Pick a patch x in M, with 
x(a,b) = p, and let v be the vector field on E 2 such that x* (v) = V. ) 

8. Prove that every surface of revolution is diffeomorphic to either a 
torus or a circular cylinder. (Similarly an augmented surface of revolu- 
tion — Ex. 12 of IV. 1 — is diffeomorphic to either a plane or a sphere.) 

9. (Cartesian products). If M and N are surfaces, let M X N be the set 
of all ordered pairs (p, q), with p in M and q in N. If x: D — ► M and 
y: E — > N are patches, let D X E be the region in E 4 consisting of 
all points (u, v, Mi, Vi) with (v, v) in D, (u x , vi) in E. Then define 
x X y: D X E ^ M X N by 

(x X y)(u, v, Mi, vi) = (x(u,v),y(ui,Vi)). 

Prove that the collection (P of all such abstract patches makes M X N 
a manifold (of dimension 4). M X AT is called the Cartesian product 
of M and N. 

The same scheme works for any two manifolds — for example, 
E 1 X E 1 is precisely E 2 . 

10. If M is an abstract surface, a proper imbedding of M in E is a one-to- 
one regular mapping F: M — > E such that the inverse function 
F~ x : F (M) — * M is continuous. Prove that the image F (M) of a proper 
imbedding is a surface in E 3 (Definition 1.2) and is diffeomorphic to M. 
If F: M — > E 3 is merely regular, then F is an immersion of M in E 
and the image F(M) is sometimes called an "immersed surface," even 
though it need not satisfy Definition 1.2. 



9 Summary 

In this chapter we have progressed from the familiar notion of surface in 
E to the general notion of manifold. Let us now reverse this process: An 
n-dimensional manifold M is a space that — near each point — is like the 
Euclidean space E n . Every manifold has a calculus consisting of differentia- 



188 CALCULUS ON A SURFACE [Chap. IV 

ble functions, tangent vectors, vector fields, mappings, and, above all, 
differential forms. The simplest manifold of dimension n is E n itself. A 
two-dimensional manifold is called a surface. Some surfaces appear in E 3 , 
some do not. This theory all comes from the usual elementary calculus on 
the best-known manifold of all, the real line. But the calculus of every 
manifold behaves in the same general way. 



CHAPTER 



V 



Shape Operators 



In Chapter II we measured the shape of a curve in E 3 by its curvature and 
torsion functions. Now we consider the analogous measurement problem 
for surfaces. It turns out that the shape of a surface M in E 3 is described 
infinitesimally by a certain linear operator £ denned on each of the tangent 
planes of M . As with curves, to say that two surfaces in E 3 have the same 
shape means simply that they are congruent. And just as with curves, we 
shall justify our infinitesimal measurements by proving that two surfaces 
with "the same" shape operators are, in fact, congruent. The algebraic 
invariants (determinant, trace, • • • ) of its shape operators thus have 
geometric meaning for the surface M. We shall investigate this matter in 
detail and find efficient ways to compute these invariants, which we test 
on a number of geometrically interesting surfaces. 

From now on the notation M <Z E 3 means a connected surface M in 
E 3 as defined in Chapter IV. 



1 The Shape Operator ofMc E 3 

Suppose that Z is a Euclidean vector field (Definition 3.7 of Chapter IV) on 
a surface M in E 3 . Although Z is defined only at points of M, the covariant 
derivative V V Z (Chapter II, Section 5) still makes sense as long as v is 
tangent to M. As usual, V V Z is the rate of change of Z in the v direction, 
and there are two main ways to compute it. 

Method 1. Let a be a curve in M that has initial velocity a (0) = v. 
Let Z a be the restriction of Z to a, that is, the vector field t — > Z(a (t) ) on 
a (Fig. 5.1). Then 

189 



190 



SHAPE OPERATORS 



[Chap. V 




V r Z 



FIG. 5.1 



V„Z = (Z a )'(0) 
where the derivative is that of Chapter II, Section 2. 

Method 2. Express Z in terms of the natural frame field of E 3 by 

z = E Zi Ui. 

Then 

v„z = EvNC/i 

where the directional derivative is that of Chapter IV, Section 3. 

It is easy to check that these two methods give consistent results. Note 
that even if Z is a tangent vector field, the co variant derivative V„Z need 
not be tangent to M. 

It follows immediately from Theorem 7.5 of Chapter IV that if M is an 
orientable surface in E 3 , then there is a unit normal vector field U on M. 
In fact, if Z is a nonvanishing normal vector field, then U = Z/ \\ Z \\ is 
still normal, and has unit length. Since M is now assumed to be connected, 
there are exactly two unit normal vector fields U and — U defined on the whole 
surface M. But even when M is not orientable, unit normals U and — U 
are still available on some neighborhood of each point of p of M (see Exer- 
cises for Chapter IV, Section 7 ) . 

We are now in a position to find a mathematical measurement of the 
shape of a surface in E . 

1.1 Definition If p is a point of M, then for each tangent vector v to M 
at p, let 

£ p (v) = -V V U 

where U is a unit normal vector field on a neighborhood of p in M. S p is 
called the shape operator of M at p (derived from C/).f (Fig. 5.2.) 

t The minus sign artificially introduced in this definition will sharply reduce the 
total number of minus signs needed later on. 



See. 1] THE SHAPE OPERATOR OF M C E 3 191 

Ufa) 




FIG. 5.2 

The tangent plane of M at any point q consists of all Euclidean vectors 
orthogonal to C/(q). Thus the rate of change V„J7 of U in the v direction 
tells how the tangent planes of M are varying in the v direction — and 
this gives an infinitesimal description of the way M itself is curving in E 3 . 

Note that if U is replaced by — U, then S p changes to — S p . 

1.2 Lemma For each point p of M CI E 3 , the shape operator is a linear 
operator 

S p : T P (M) -+ T P (M) 

on the tangent plane of M at p. 

Proof. In Definition 1.1, U is a unit vector field, so U'U = 1. Thus by 
a Leibnizian property of covariant derivatives, 

= y[U*U\ = 2V V U-U(p) = -2£ p (v).tf(p) 

where v is tangent to M at p. Since U is also a normal vector field, it follows 
that S p (v) is tangent to M at p. Thus S P is a function from T P (M) to 
T P (M). (It is to emphasize this that we use the term "operator" instead of 
' 'transformation. ' ' ) 

The linearity of S p is a consequence of a linearity property of covariant 
derivatives. 

S p (av + 6w) = —Vav+bwU = - (aV v U + bVJU) 

= aS p (y) + bS p (w). | 

At each point p of M a E there are actually two shape operators ±S P 
derived from the two unit normals ±U near p. We shall refer to all of 
these, collectively, as the shape operator S of M. Thus if a choice of unit 
normal is not specified, there is a (relatively harmless) ambiguity of sign. 

1.3 Example Shape operators of some surfaces in E 3 . (1) Let S be 
the sphere of radius r consisting of all points p of E 3 with || p || = r. Let 
U be the "outward normal" on S. Now as U moves away from any point 
p in the direction v, evidently U topples forward in the exact direction of 
v itself (Fig. 5.3). Thus S(\) must have the form — cv. 

In fact, using gradients as in Example 3.9 of Chapter IV, we find 



192 



SHAPE OPERATORS 



[Chap. V 




C/ = iZ 






But then 



V V U =-EvW^-(p) = 



Thus S(\) = — v/r for all v. So the shape 
operator S is merely scalar multiplication by 
— 1/r. This uniformity in S reflects the round- 
fig. 5.3 ness of spheres: They bend the same way in all 

directions at all points. 

(2) Let P be a plane in E 3 . A unit normal vector field U on P is evidently 
parallel in E (constant Euclidean coordinates) (Fig. 5.4). Hence 

S(v) = -V V U = 

for all tangent vectors v to P. Thus the shape operator is identically zero 
— to be expected, since planes do not bend at all. 

(3) Let C be the circular cylinder x + y 2 = r 2 in E 3 . At any point p 
of C, let ei and e 2 be unit tangent vectors, with e x tangent to the ruling of 
the cylinder through p, and ^ tangent to the cross-sectional circle. 

Use the outward normal U as indicated in Fig. 5.5. 

Now, when U moves from p in the ei direction, it stays parallel to itself 
just as on a plane; hence *S(d) = 0. When U moves in the e 2 direction, it 
topples forward exactly as on a sphere of radius r; hence *S(e 2 ) = — e 2 /r. 
In this way £ describes the "half -flat, half-round" shape of a cylinder. 

(4) The saddle surface M: z = xy. For the moment we investigate S 
only at p = (0, 0, 0) in M . Since the x and y axes of E 3 lie in M , the vectors 
iii = (1,0,0) and u 2 = (0, 1,0) are tangent to M at p. We use the "up- 
ward" unit normal U, which at p is (0,0, 1). Along the x axis, U stays 



U 




U(p) 



FIG. 5.4 



FIG. 5.5 



Sec. 1] 



THE SHAPE OPERATOR OFMCP 



193 




FIG. 5.6 



orthogonal to the x axis, and as we proceed in the ui direction, U swings 
from left to right (Fig. 5.6). In fact, a routine computation (Exercise 3) 
shows that V Ul U = -u 2 . Similarly we find V U2 U = — u x . 
Thus the shape operator of M at p is given by the formula 

S(aui + 6u 2 ) = b\ii + au 2 . 

These examples clarify the analogy between the shape operator of a 
surface and the curvature and torsion of a curve. In the case of a curve, 
there is only one direction to move, and k and t measure the rate of change 
of the unit vector fields T and B (hence N). For a surface only one unit 
vector field is intrinsically determined — the unit normal U. Furthermore, 
at each point, there are now a whole plane of directions in which U can 
move, so that rates of change of U are measured, not numerically, but by 
the linear operators S. 

1.4 Lemma For each point p of I c E 3 the shape operator 

S: T P M -> T P M 
is a symmetric linear operator; that is, 

/S(v)»w = S(w)'\ 
for any pair of tangent vectors to M at p. 

We postpone the proof of this crucial fact to Section 4, where it occurs 
naturally in the course of general computations. 

From the viewpoint of linear algebra, a symmetric linear operator on a 
two-dimensional vector space is a very simple object indeed. For a shape 
operator, its characteristic values and vectors, its trace and determinant, 
all turn out to have geometric meaning of first importance for the surface 
M a E 3 . 



194 SHAPE OPERATORS [Chap. V 

EXERCISES 

1 . Let a be a curve in M (Z E 3 . If U is a unit normal of M restricted to 
the -curve a, show that S(a') = — V . 

2. Consider the surface M: z = f(x,y), where 

/(0,0) = /,(0,0) = /,(0,0) = 0. 

(The subscripts indicate partial derivatives.) Show that 

(a) The vectors ui = Ui(0) and u 2 = ^(O) are tangent to M at the 
origin 0, and 

~ UUl - fyU2 + Us 

Vi + U + /„» 

is a unit normal vector field on M. 

(b) £( Ul ) =/«(0,0)u 1 +/x„(0,0)ua 
^(U2) =/,x(0,0) Ul +/ yw (0,0)u 2 . 

(Note: The square root in the denominator is no real problem here be- 
cause of the special character of/ at (0, 0). In general, direct computa- 
tion of S is difficult, and in Section 4 we shall establish indirect ways of 
getting at it.) 

3. (Continuation) . In each case, express S(aui + 6u 2 ) in terms of ui and 
u 2 , and determine the rank of *S at (rank S is dimension of image S: 
0,1, or 2). 

(a) z = xy. (c) z = (x + yf. 

(b) z = 2x 2 + y. (d) z = xy. 

4. Let M be a surface in E oriented by a unit normal vector field 

U = giUi + QiVi + gzUz. 

Then the Gauss mapping G: M —* 2 of M sends each point p to the 
point (<7i(p), <72(p)> <73 (p)) of the unit sphere 2. Pictorially: Move 
U(p) to the origin by parallel motion; there it points to G(p) (Fig. 5.7). 



U(p) 




FIG. 5.7 



Sec. 2] NORMAL CURVATURE 195 

Thus G completely describes the turning of U as it traverses M. 

For each of the following surfaces, describe the image G(M) of the 
Gauss mapping in the sphere 2 (use either normal) : 

(a) Cylinder, x + y 2 = r . 

(b) Cone, z = y/x* + y 2 . 

(c) Plane, x + y + z = 0. 

(d) Sphere, (x - l) 2 + y 2 + (z + 2) 2 = 1. 

5. Let G: T — > 2 be the Gauss mapping of the torus T (as in IV. 2.6) 
derived from its outward unit normal U. What are the image curves 
under G of the meridians and parallels of T? Which points of 2 are the 
image of exactly two points of T? 

6. Let G: M — > 2 be the Gauss mapping of the saddle surface M : z = xy 
derived from the unit normal U obtained as in Exercise 2. What is the 
image under G of one of the straight lines, y constant, in M? How much 
of the sphere is covered by the entire image G(M) ? 

7. Show that the shape operator of M is (minus) the derivative of its 
Gauss mapping: If S and G: M -> 2 both derive from U, then S(\) 
and — (j* (v) are parallel for every tangent vector v to M. 

8. An orientable surface has two Gauss mappings derived from its two 
unit normals. Show that they differ only by the antipodal mapping of 
2 (Ex. 5 of IV. 5). Define a Gauss-type mapping for a nonorientable 
surface in E . 

9. If V is a tangent vector field on M (with unit normal U), then by the 
pointwise principle, S(V) is the tangent vector field on M whose value 
at each point p is S p (V(p)). Show that 

S(V)-W = VyW-U. 

Deduce that the symmetry of S is equivalent to the assertion that 
the bracket 

[V, W] = v v w - v w v 

of two tangent vector fields is again a tangent vector field. 



2 Normal Curvature 

Throughout this section we shall work in a region of M CI E which has been 
oriented by the choice of a unit normal vector field U, and we use the shape 
operator S derived from U. 

The shape of a surface in E 3 influences the shape of the curves in M. 



196 



SHAPE OPERATORS 



[Chap. V 




FIG. 5.8 

2.1 Lemma If a is a curve inMc E 3 , then 

a ".U = S(a')-a. 

Proof. Since a is in M, its velocity a is always tangent to M . Thus 
a • U = 0, where as in Section 1 we restrict U to the curve a. Differentiation 
yields 

<*".[/ + d'Xf = 0. 

But from Section 1, we know that S(a ) = — U. Hence 

a ».U = -if. a ' = S(d).d. | 

Geometric interpretation: at each point a" ' *U is the component of accel- 
eration a" normal to the surface M (Fig. 5.8). The lemma shows that this 
component depends only on the velocity a and the shape operator of M. 
Thus all curves in M with a given velocity v at point p will have the same nor- 
mal component of acceleration at p, namely, $(v)«v. This is the component 
of acceleration which the bending ofMinJZ forces them to have. 

Thus if we standardize v by reducing it to a unit vector u, we get a 
measurement of the way M is bent in the u direction. 

2.2 Definition Let u be a unit vector tangent toMcE at a point p. 
Then the number k(u) = $(u)»uis called the normal curvature of M in 
the u direction. 

To make the term direction precise, we define a tangent direction to M at 
p to be a one-dimensional subspace L of T P (M), that is, a line through the 
zero vector (located for intuitive purposes at p) (Fig. 5.9). Any nonzero 
tangent vector at p determines a direction L, but we prefer to use one of 



FIG. 5.9 



Sec. 2] 



NORMAL CURVATURE 



197 




FIG. 5.10 



the two unit vectors ±u in L. Note that 

fc(u) = S(u)-u = S(-u).(-u) = fc(-u). 

Thus, although we evaluate k on unit vectors, it is, in effect, a real-valued 
function on the set of all tangent directions to M. 

Given a unit tangent vector to M at p, let a be a (unit speed) curve in 
M with initial velocity a'(0) = u. Using the Frenet apparatus of a, the 
preceding lemma gives 

fc(u) = S(u)-u = a"(0).£/(p) = K(0)iV(0).t/(p) 

= k(0) cost?. 

Thus the normal curvature of M in the u direction is k(0) cos #, where 
k(0) is the curvature of a at a (0) = p, and # is the angle between the 
principal normal N(0) and the surface normal U(p), as in Fig. 5.10. 

Given u, there is a natural way to choose the curve so that # is or v. 
In fact, if P is the plane determined by u and C/(p), then P cuts from M 
(near p) a curve <r called the normal section of M in the u direction. If we 
give a unit-speed parametrization with <r (0) = u, it is easy to see that 
N(0) = dbU(p). (er"(0) = k(0)N(0) is orthogonal to </(0) = u and 
tangent to P.) Thus for a normal section in the u direction (Fig. 5.11), 

fc(u) = K a (0)N(0)-U(p) = ±k„(0). 

Thus it is possible to make a reasonable estimate of the normal curva- 
tures in various directions on a surface M CI E 3 by picturing what the cor- 
responding normal sections would look like. We know that the principal 
normal N of a curve tells in which direction it is turning. Thus the preceding 
discussion gives geometric meaning to the sign of the normal curvature 
k(u) (relative to our fixed choice of U). 



198 



SHAPE OPERATORS 



[Chap. V 




FIG. 5.11 



(1) If fc(u) > 0, then AT (0) = U(p), so the normal section a is bending 
toward U(p) at p (Fig. 5.12). Thus in the u direction the surface M is 
bending toward U(p). 



U(p) 



N(0) = CT(p) 



FIG. 5.12 



N(0) 
FIG. 5.13 



(2) If k(u) < 0, then JV(0) = — U(p), so the normal section a is 
bending away from U(p) at p. Thus in the u direction M is bending away 
from U(p) (Fig. 5.13). 

(3) If k(u) = 0, then k,(0) = (iV(0) is undefined). Here the normal 
section a is not turning at <r(0) = p. We cannot conclude that in the u 
direction M is not bending at all, since k might be zero only at o-(O) = p. 
But we can conclude that its rate of bending is unusually small. 

In different directions at a fixed point p, the surface may bend in quite 
different ways. For example, consider the saddle surface z = xy in Example 
1.3. If we identify the tangent plane of M at p = (0, 0, 0) with the xy plane 
of E 3 , then clearly the normal curvature in the direction of the x and y axes 
is zero, since the normal sections are straight lines. However, Fig. 5.6 shows 
that in the tangent direction given by the line y = a;, the normal curva- 
ture is positive, for the normal section is a parabola bending upward. 
(C/(p) = (0,0,1) is "upward.") But in the direction of the line y = —x, 
normal curvature is negative, since this parabola bends downward. 



Sec. 2] 



NORMAL CURVATURE 

U(p) 



199 



E* 



(^j2^^« 




T,(M\ 



'•* •■•''■*- **■•?& V 




FIG. 5.14 

Let us now fix a point p of M C E 3 and imagine that a unit tangent 
vector u at p revolves, sweeping out the unit circle in the tangent plane 
T P (M). From the corresponding normal sections, we get a moving picture 
of the way M is bending in every direction at p (Fig. 5.14). 

2.3 Definition Let p be a point of M CZ E 3 . The maximum and minimum 
values of the normal curvature A;(u) of M at p are called the principal 
curvatures of M at p, and are denoted by fa and fa. The directions in which 
these extreme values occur are called principal directions of M at p. Unit 
vectors in these directions are called principal vectors of M at p. 

Using the normal-section scheme discussed above, it is often fairly easy 
to pick out the directions of maximum and minimum bending. For example, 
if we use the outward normal (U) on a circular cylinder C, then the normal 
sections of all bend away from U, so k (u) ^ 0. Furthermore, it is reasonably 
clear that the maximum value fa = occurs only in the direction of a rul- 
ing, minimum value fa < occurs only in the direction tangent to a cross 
section, as in Fig. 5.15. 




FIG. 5.15 



200 SHAPE OPERATORS [Chap. V 

An interesting special case occurs at points p for which fa = fa. The 
maximum and minimum normal curvature being equal, it follows that k (u) 
is constant: M bends the same amount in all directions at p (and all direc- 
tions are principal). 

2.4 Definition A point p of M C E 3 is umbilic provided the normal 
curvature A;(u) is constant on all unit tangent vectors u at p. 

For example, what we found in (1 ) of Example 1.3 was that every point 
of the sphere 2 is umbilic, with fa = fa = — 1/r. 

2.5 Theorem (1) If p is an umbilic point of M C E 3 , then the shape 
operator S at p is just scalar multiplication by k = fa = fa. 

(2 ) If p is a nonumbilic point, fa ?* fa, then there are exactly two prin- 
cipal directions, and these are orthogonal. Furthermore, if d and e2 are 
principal vectors in these directions, then 

S(ei) = fae! /S(e 2 ) = A^d- 

In short, the principal curvatures of M at p are the characteristic values 
of S, and the principal vectors of M at p are the characteristic vectors of S. 

Proof. Suppose that k takes on its maximum value fa at d, so 

fa = k(ei) = Sid) »ei. 

Let C2 be merely a unit tangent vector orthogonal to ei (presently we shall 
show that it is also a principal vector. ) 

If u is any unit tangent vector at p, we write 

u = u(#) = cei + se2 

where c = cos ■&, s = sin & (Fig. 5.16). Thus normal curvature k at p be- 
comes a function on the real line: k(&) = k(u(&)). 

For 1 ^ i, j <^ 2, let Sa be the number S(ei)*ej. Note that S n = fa, and 
by the symmetry of the shape operator, $ 12 = S^. We compute 

&(#) = »S(cei + se 2 )«(cei + se 2 ) 

(1) 
= cS n + 2scSu + s 2 £ 22 . 



Hence 



^ (*) = 2 SC (£ 22 - Sn) + 2(c 2 - s 2 )S 12 . (2) 

d& 



If # = 0, then c = 1 and s = 0, so u(0) = d. Thus, by assumption, 
fc(#) is a maximum at & = 0, so (dk/d&)(0) = 0. It follows immediately 
from (2) that S n = 0. 

Since d, e 2 is an orthonormal basis for T P (M), we deduce by orthonormal 



Sec. 2] 



NORMAL CURVATURE 



201 



expansion that 



£(ei) = S u ei /S(e 2 ) = &2e 2 . 



(3) 



Now if p is umbilic, then S 22 = fcfe) is the same as S n = &(ei) = h, so 
(3) shows that S is scalar multiplication by h = k 2 . 
If p is not umbilic, we look back at (1), which has become 

fc(#) = c % + s 2 &2. (4) 

Since h is the maximum value of fc(#), and fc(t^) is now nonconstant, it 
follows that h > S*. But then (4) shows: (a) the maximum value h is 
taken on only when c = ±1, s = 0, that is, in the d direction, and (b) the 
minimum value k 2 is £22, and is taken on only when c = 0, s = ±1, that is, 
in the e 2 direction. This proves the second assertion in the theorem, since 
(3 ) now reads : 

£(ei) = hd, SM = k^. I 

Contained in the preceding proof is Euler's formula for the normal curva- 
ture of M in all directions at p. 

2.6 Corollary Let k u k 2 and e u d be the principal curvatures and vectors 
oflcE 8 at p. Then if u = cos tfei + sin t?e 2 , the normal curvature of 
M in the u direction is (Fig. 5.16) 

k(u) = h cos 2 # + kv sin 2 t?. 

Here is another way to show how the principal curvatures h and fc 2 
control the shape of M near an arbitrary point p. Since the position of M 
in E 3 is of no importance, we can assume that (1) p is at the origin of E , 
(2) the tangent plane f p (M) is the xy plane of E 3 , and (3) the x and y axes 
are the principal directions. Near p, M can be expressed as M: z = f(x,y), 
as shown in Fig. 5.17, and the idea is to construct an approximation of M 
near p by using only terms up to quadratic in the Taylor expansion of the 




(x,y,f(x,y) 



T P (M) 




FIG. 5.16 



FIG. 5.17 



202 SHAPE OPERATORS [Chap. V 

function/. Now (1) and (2) imply/ = f x = f y ° = 0, where the subscripts 
indicate partial derivatives, and the superscript zero denotes evaluation at 
x = 0, y = 0. Thus the quadratic approximation of / near (0, 0) reduces to 

f(x,y) ~ h(f xx x + 2f xy xy + f yy y 2 ). 

In Exercise 2 of Section 1 we found that for the tangent vectors 

ux = (1,0,0) and u 2 = (0,1,0) 

at p = 

£(ui) = -V Ml C7 = f xx jx 1 +/° J/ u 2 

S(U 2 ) = —V U2 U= f xy U! + flyU 2 . 

By condition (3) above, ui and u 2 are principal vectors, so it follows 
from Theorem 2.5 that h = f xx , k 2 = f yy , and /J, = 0. 

Substituting these values in the quadratic approximation of /, we con- 
clude that the shape of M near p is approximately the same as that of the 
surface 

M:z = \{hx 2 + k 2 y 2 ) 

near 0. M is called the quadratic approximation of M near p. It is an ana- 
logue for surfaces of a Frenet approximation of a curve. 

From Definition 2.2 through Corollary 2.6 we have been concerned with 
the geometry of M C E 3 near one of its points p. These results thus apply 
simultaneously to all the points of the oriented region on which, by our 
initial assumption, the unit normal U is defined. In particular then, we 
have actually defined principal curvature functions ki and k 2 on 0. At each 
point p of 0, &i(p) and A; 2 (p) are the principal curvatures of M at p. We 
emphasize that these functions are only defined "modulo sign": If U is 
replaced by —U, they become — h and — k 2 . 



EXERCISES 

1. Use the results of Example 1.3 to find the principal curvatures and 
principal vectors of 

(a) The cylinder, at every point. 

(b) The saddle surface, at the origin. 

2. If v is a nonzero tangent vector (not necessarily of unit length), show 
that the normal curvature of M in the direction determined by v is 

k(v) = S(\)'\/y\. 

3. For each integer n ^ 2, let a n be the curve t — > (r cos t, r sin t, ±t n ) 



Sec. 3] GAUSSIAN CURVATURE 203 

in the cylinder M: x + y 2 = r 2 . These curves all have the same velocity 
at t = 0; test Lemma 2.1 by showing that they all have the same normal 
component of acceleration at t = 0. 

4. For each of the following surfaces, find the quadratic approximation 
near the origin: 

(a) z = exp(z 2 + y 2 ) - 1. 

(b) z = log cos x — log cos y. 

(c) z = (x + Zy)\ 

5. Justify the first sentence in the proof of Theorem 2.5 : Show that k has a 
maximum value. 



3 Gaussian Curvature 

In the preceding section we found the geometrical meaning of the charac- 
teristic values and vectors of the shape operator. Now we examine the de- 
terminant and trace of S. 

3.1 Definition The Gaussian curvature ofMcE is the real- valued func- 
tion K = det S on M. Explicitly, for each point p of M, the Gaussian curva- 
ture K(p) of M at p is the determinant of the shape operator S of M at p. 

The mean curvature of M c E 3 is the function H = \ trace S. Gaussian 
and mean curvature are expressed in terms of principal curvature by 

3.2 Lemma K = kfa, H = (h + k 2 )/2. 

Proof. The determinant (and trace) of a linear operator may be defined 
as the common value of the determinant (and trace) of all its matrices. 
If e x and 62 are principal vectors at a point p, then by Theorem 2.5, we have 
>S(ei) = fci(p)ei and /S(e 2 ) = fc 2 (p)e2. Thus the matrix of S at p with re- 
spect to d, e 2 is 

'fcx(p) 

^0 fc 2 (p), 

This immediately gives the required result. | 

A significant fact about the Gaussian curvature: It is independent of the 
choice of the unit normal U. If U is changed to — U, then the signs of both 
ki and fc 2 change, so K = kik 2 is unaffected. This is obviously not the case 
with mean curvature H = (h -{- k 2 )/2, which has the same ambiguity of 
sign as the principal curvatures themselves. 

The normal section method in Section 2 lets us tell, by inspection, ap- 
proximately what the principal curvatures of M are at each point. Thus we 



204 



SHAPE OPERATORS 



[Chap. V 



U(p) 




hip) < 
hip) < 



U(p) 




FIG. 5.18 



FIG. 5.19 



get a reasonable idea of what the Gaussian curvature K = hh is at each 
point p by merely looking at the surface M. In particular, we can usually 
tell what the sign of Kip) is — and this sign has an important geometric 
meaning, which we now illustrate. 

3.3 Remark The sign of Gaussian curvature at a point p. 

(1) Positive. If Kip) > 0, then by Lemma 3.2, the principal curvatures 
hip) and A; 2 (p) have the same sign. By Corollary 2.6, either k (u) > for 
all unit vectors u at p or k(u) < 0. Thus M is bending away from its 
tangent plane T P (M) in all tangent directions at p. (Fig. 5.18). 

The quadratic approximation of M near p is the paraboloid 

2z = hip)x 2 + hip)y\ 

(2) Negative. If Kip) < 0, then by Lemma 3.2 the principal curvatures 
h (p) and k 2 (p) have opposite signs. Thus the quadratic approximation of 
M near p is a hyperboloid, so M also is saddle-shaped near p (Fig. 5.19). 

(3) Zero. If K(p) = 0, then by Lemma 3.2 there are two cases: 

(a) Only one principal curvature is zero, say 

h(p) * 0, A; 2 (p) = 0. 

(b) Both principal curvatures are zero: 

hip) = fc 2 ( P ) = 0. 

In case (a) the quadratic approximation is the cylinder 2z = hip)x 2 , so 
M is trough-shaped near p (Fig. 5.20). 

In case (b), the quadratic approximation reduces simply to the plane 
z = 0, so we get no information about the shape of M near p. 

A torus of revolution T provides a good example of these different cases. 
At points on the outer half of T, the torus bends away from its tangent 
plane as one can see from Fig. 5.21 ; hence K > on 0. But near each point 
p of the inner half d, T is saddle-shaped and cuts through T P iM). Hence 
K < on d. 



Sec. 3] 



GAUSSIAN CURVATURE 



205 



U(P) 




*i(p) > 
M l hip) = 



FIG. 5.20 




FIG. 5.22 

Near each point on the two circles (top and bottom) which separate 
and #, the torus is trough-shaped; hence K = there. (A quantitative 
check of these qualitative results is given in Section 6.) 

In case 3 (b ) above, where both principal curvatures vanish, p is called a 
planar point of M . (There are no planar points on the torus.) For example, 
the central point p of a monkey saddle, say 

M: z = x(x + VSy) (x - VSy), 

is planar. Here three hills and valleys meet, as shown in Fig. 5.22. Thus 
p must be a planar point — the shape of M near p is too complicated for the 
other three possibilities in Remark 3.3. 

We consider now some ways to compute Gaussian and mean curvature. 

3.4 Lemma If v and w are linearly independent tangent vectors at a 
point p of M cz E 3 , then 

S(v) X S(w) = K(p)v X w 

S(v) X w + v X S(w) = 2H(p)v X w. 

Proof. Since v, w is a basis for the tangent plane T P (M), we can write 

iS(v) = av + 6w 

S(w) = cv + dw. 
Thus 



206 



SHAPE OPERATORS 



[Chap. V 



a b 
c d 



is the matrix of S with respect to the basis v, w. Hence 

K(p) = det S = ad - be H(p) = \ trace S = |(o + d). 
Using standard properties of the cross product, we compute 
S(y) X S(w) = (ax + 6w) X (cv + dw) 

= (ad - be) v X w = K(p) v X w 
and a similar calculation gives the formula for H(p). 



I 



Thus if V and W are tangent vector fields that are linearly independent 
at each point of an oriented region, we have vector field equations 

S(V) X S(W) = KV XW 

S(V) XW +V X S(W) =2HV XW. 

These may be solved for K and H by dotting each side with the normal 
vector field V XW, and using the Lagrange identity (Exercise 6 ) . We then 
find 



K = 



SV-V SV-W 
SW-V SW-W 



H = 



V*V V'W 
W-V W'W 



SV-V sv 
W-V w- 


•W 
W 


+ 


V'V V'W 
SW-V SW-W 


2 


W- 


V 
V 


V'W 
W'W 





(The denominators are never zero, since the independence of V and W is 
equivalent to (V X W)'(V X W) > 0.) In particular the functions K 
and H are differ entiable. 

Once K and H are known, it is a simple matter to find ki and k 2 . 

3.5 Corollary On an oriented region in M, the principal curvature 
functions are 

k h k 2 = H ± VH 2 - K. 

Proof. To verify the formula it suffices to substitute 

K = A;^ and H = (h + A*)/2, 

and note that 



H 2 - K = 



(a* + k 2 y 



— kik 2 



(fa — faf 



Sec. 3] GAUSSIAN CURVATURE 207 

A more enlightening derivation (Exercise 4) uses the characteristic poly- 
nomial of S. 

This formula shows only that ki and k 2 are continuous functions on 0; 
they need not be differentiable since the square-root function is badly be- 
haved at zero. The identity in the proof shows that H 2 — K is zero only at 
umbilic points, however, so ki and k 2 are differentiable on any oriented region 
free of umbilics. 

A natural way to single out special types of surfaces in E 3 is by restric- 
tions on Gaussian and mean curvature. 

3.6 Definition A surface M in E 3 is flat provided its Gaussian curvature 
is zero, and minimal provided its mean curvature is zero. 

As expected, a plane is flat, for by Example 1.3 its shape operators are 
all zero, so K = det S = O.\0n a circular cylinder, (3 ) of Example 1.3 shows 
that S is singular at each point p, that is, has rank less than the dimension 
of the tangent plane T P (M). Thus, although S itself is never zero, its de- 
terminant is always zero, so cylinders are also flat. This terminology seems 
odd at first for a surface so obviously curved, but it will be amply justified 
in later work. 

Note that minimal surfaces have Gaussian curvature K ^ 0, because 
if H = (fci + k 2 )/2 = 0, then h = -k 2 , so K = hk 2 ^ 0. 

Another notable class of surfaces consists of those with constant Gaussian 
curvature. As mentioned earlier, Example 1.3 shows that a sphere of radius 
r has ki = k 2 = —1/r (for U outward). Thus the sphere 2 has constant 
positive curvature K = 1/r 2 : The smaller the sphere, the larger its curvature. 

We shall find many examples of these various special types of surface 
as we proceed through this chapter. 



EXERCISES 

1. Show that there are no umbilics on a surface with K < 0, and that if 
K ^ 0, umbilic points are planar. 

2. Let iii and u 2 be orthonormal tangent vectors at a point p of M. What 
geometric information can be deduced from each of the following condi- 
tions on S at p? 

(a) 5(u!).u s = 0. (c) S( Ul ) X #(u 2 ) = 0. 

(b) 5(u0 + S(u 2 ) = 0. (d) S( Ul )-S(u 2 ) = 0. 

3. (Mean curvature). Prove that 

(a) the average value of the normal curvature in any two orthogonal 
directions at p is H(p). (The analogue for K is false.) 



208 SHAPE OPERATORS [Chap. V 

(b) H(p) = (1/2*) f **(*) dfi, 

where k(&) is the normal curvature, as in Corollary 2.6. 

4. The characteristic polynomial of an arbitrary linear operator S is 

p(k) = det(A - kl), 

where A is any matrix of S. 

(a) Show that the characteristic polynomial of the shape operator is 
k 2 - 2Hk + K. 

(b) Every linear operator satisfies its characteristic equation; that is, 
p(S) is the zero operator when S is formally substituted in p(k). 
Prove this in the case of the shape operator by showing that 

SvSw - 2HSvw + Kvw = 

for any pair of tangent vectors to M . 
The real-valued functions 

I(v,w) = vw, II(v,w) = Svw, 

and 

III(v,w) = S vw = SvSw, 

defined for all pairs of tangent vectors to an oriented surface, are 
traditionally called the first, second, and third fundamental forms of M. 
They are not differential forms; in fact, they are symmetric in v and w 
rather than alternate. The shape operator does not appear explicitly in 
the classical treatment of this subject; it is replaced by the second fun- 
damental form. 

5. (Dupin curves). For a point p of an oriented region of M, let C be the 
intersection of M near p with its tangent plane f p (M); specifically, C 
consists of those points of M near p which lie in the plane through p 
orthogonal to U(p). C may be approximated by substituting for M its 
quadratic approximation M; thus Co is approximated by the curve 

C : hx 2 + k*y 2 = 0, near (0, 0). 

(a) Describe Co in each of the three cases K(p) > 0, K(p) < 0, and 
K(p) = (not planar). 

(b) Repeat (a) with Co replaced by C e and C_e, where the tangent 
plane has been replaced by the two parallel planes at distance ±e 
from it. 

(c) This scheme fails for planar points since the quadratic approxima- 
tion becomes M : z = 0. For the monkey saddle, sketch C , Ce, and 
C_e. 



Sec 3] 



GAUSSIAN CURVATURE 



209 




FIG. 5.23 

6. If v, w, a, and b are vectors in E , prove the Lagrange identity 

v»a v»b 



(v X w).(a Xb) = 



w»a w»b 



7. (Parallel surfaces). Let M be a surface oriented by U; for a fixed number 
e (positive or negative) let F: M — ► E 3 be the mapping such that 

F(p) = p + eC/(p). 

(a) If v is tangent to M at p, show that v = F* (v) is v — eS(\). De- 
duce that 



where 



v X w = «/(p) v X w, 



J = 1 - 2etf + e 2 K = (1 - efci)(l - ek 2 ). 



If the function «/ does not vanish on M (M is compact and | e | small), 
this shows that F is a regular mapping, so the image 

M = F(M) 

is at least an immersed surface in E 3 (Ex. 10 of IV.8) . M is said to 
be -parallel to M at distance e (Fig. 5.23). 

(b) Show that the canonical isomorphisms of E make U a unit normal 
on M for which S(r) = S(\). 

(c) Derive the following formulas for the Gaussian and mean curvatures 
of M: 

K(F) = K/J; S(F) = (H - eK)/J. 

8. (Continuation) 

(a) Check the results in (c) in the case of a sphere of radius r oriented 
by the outward normal U. Describe the mapping F = Fe when e 
is 0, —r, and — 2r. 



210 SHAPE OPERATORS [Chap. V 

(b) Starting from an orientable surface with constant positive Gaussian 
curvature, construct a surface with constant mean curvature. 



4 Computational Techniques 

We have denned the shape operator S of a surface M in E and found 
geometrical meaning for its main algebraic invariants: Gaussian curvature 
K, mean curvature H, principal curvatures fci and k 2 , and (at each point) 
principal vectors ei and e 2 . We shall now see how to express these invariants 
in terms of patches in M . 

If x: D — * M is a patch inlc E 3 , we have already used the three real- 
valued functions 

E = X U »X U , F = X U *X V = x v *x u , G = x„«x„. 

on D. Here E > and G > are the squares of the speeds of the u- and 
y-parameter curves of x, and F measures the coordinate angle # between 
x u and x„, since 

F = x u »x v — || x u || || x„ || cos # = s/EG cos &. 

(Fig. 5.24). E, F, and G are the "warping functions" of the patch x: They 
measure the way x distorts the flat region D in E 2 in order to apply it to 
the curved region x(D) in M. These functions completely determine the 
dot product of tangent vectors at points of x(D), for if 

v = Vix u + v 2 x v and w = Wix u + w 2 x u , 

then 

vw = EviWi + F{viw 2 + v 2 Wi) + Gv 2 w 2 . 

(In such equations we understand that x„, x„, E, F, and G are evaluated at 
(u, v) where x(u,v) is the point of application of v and w.) 

Now x M X x„ is a function on D whose value at each point (u, v) of D 
is a vector orthogonal to both x u (u, v) and x v (u, v) — and hence normal to 
M at x(u, v). Furthermore, by Lemma 1.8 of Chapter II, 

|| Xu X x„ || 2 = EG- F\ 






FIG. 5.24 



Sec. 4] COMPUTATIONAL TECHNIQUES 211 

Since x is, by definition, regular, this real-valued function on D is never 
zero. Thus we can construct the unit normal function 

j j X u y\ X v 



Xi4 P\ X-p 



on D, which assigns to each (u, v) in D a unit normal vector at x(u, v). We 
emphasize that in this context, U, like x u and x„, is not a vector field on 
x(D), but merely a vector- valued function on D. Nevertheless we may 
regard the system x u , x v , U as a kind of defective frame field. At least U 
has unit length and is orthogonal to both x u and x„, even though x„ and 
x v are generally not orthonormal. 

In this context, covariant derivatives are usually computed along the 
parameter curves of x, where by the discussion in Section 1, they reduce to 
partial differentiation with respect to u and v. As in the case of x u and x„, 
these partial derivative are again denoted by subscripts u and v. If 

x{u, v) = (xi(u, v), x 2 (u, v), Xz(u, v)), 

then just as for x„ and x„ on page 134, we have 

(d 2 Xi d X2 d Xz\ 

_ ( d 2 £i d 2 X2 d Xz \ 
Xvu ~ \dudv ' dudv ' dudv/x 

(d 2 Xi d 2 X2 d 2 X 3 \ 

Evidently x uu and x vv give the accelerations of the u- and ^-parameter 
curves. Since order of partial differentiation is immaterial, x uv = x„„, which 
gives both the covariant derivative of x M in the x„ direction, and x„ in the 
x u direction. 

Now if 8 is the shape operator derived from U, we define three more 
real-valued functions on D: 

( = S(x u )»x u 

m = S(x u )*X v = S(x v )*X u 
n = S(X V )'X V . 

Because x u , x„ gives a basis for the tangent space of M at each point of 
x(D), it is clear that these functions uniquely determine the shape operator. 
Since this basis is generally not orthonormal, f, m, and n do not lead to 
simple expression for S(x u ) and *S(x„) in terms of x u and x„. In the formulas 
preceding Corollary 3.5, however, they do provide simple expressions for 
Gaussian and mean curvature. 



212 



SHAPE OPERATORS 



[Chap. V 



4.1 Corollary If x is a patch in M CZ E , then 



K(x) = 



L - 



EG - F 2 ' 



H(x) = 



Gl + En - 2F„ 

2(EG - F*) 



Proof. At a point p of x(D), the formulas on page 206 express K(p) and 
H(p) in terms of tangent vectors V(p) and W(p) at p. If V(p) and W(p) 
are replaced by the tangent vectors x u (u, v) and x v (u, v) at x(u, v), we 
find the required formulas for K(x(u, v)) and H(x(u, v)). I 

When the patch x is clear from context, we shall usually abbreviate the 
composite functions K(x) and H(x) to merely K and H. 

By a device like that used in Lemma 2.1 we can find a simple way to 
compute I, m , and n — and thereby K and i7. For example, since U*x u = 0, 
partial differentiation with respect to v — that is, ordinary differentiation 
along v-parameter curves — yields 

= — (U'x u ) = U v *x u + U'x uv . 
dv 

(Recall that U v is the covariant derivative of the vector field v — > U(u , v) 
on each y-parameter curve, u = uo.) Since x v gives the velocity vectors 
of such curves, Exercise 1 of Section 1 shows that U v = —S(x v ). Thus the 
preceding equation becomes 

S(x v )-x u = U'x uv . 

(Fig. 5.25). Three similar equations may be found by replacing u by v, 
and v by u. In particular, 

S(x u )'X v = U'X VU = U*x uv = S(x v )*x u . 







FIG. 5.25 



Sec. 4] 



COMPUTATIONAL TECHNIQUES 



213 




FIG. 5.26 



Again, since x„ and x„ give a basis for the tangent space at each point, this 
is sufficient to prove that S is symmetric (Lemma 1.4). 

4.2 Lemma If x is a patch inlc E 3 , then 

i = S(x u )'X u = U'Xuu 
m = S(x u )»X v = U*X UV 
n = S(x v )*X v = U'X VV . 

The first equation in each case is just definition, and u and v may be 
reversed in the formulas for m . 

4.3 Example Computation of Gaussian and mean curvature 

(1) He/i'co/d (Exercise 7 of Section 2, Chapter IV). This surface H, 
shown in Fig. 5.26, is covered by a single patch 



x(u, v) — {u cos v, u sin v, bv), 6^0, 



for which 



x M = (cos v, sin v, 0) 

x„ = ( — u sin v, u cos v, b) 



E = 1 

F = 

G = b 2 + u\ 



Hence 



x„ X x„ = (b sin v, —b cos v, u). 



214 SHAPE OPERATORS [Chap. V 

To find K alone it is not necessary to compute E, F, and G, but it is 
wise to do so anyway, since the identity 

1 x u Xx,|= VEG - F 2 

then gives a check on the length of x u X x„. (Its direction may also be 
checked, since it must be orthogonal to both x u and x„.) If we denote 
|| x u X x„ || by W, then W = -\/b 2 + u 2 for the helicoid, so the unit normal 
function is 

TJ _ x M X x„ (6 sin v, — b cos v, u) 



w v& 2 + 



Next we find 



x«« u 

x u » = ( — sin v, cos v, 0) 

x„„ = (— u cos v, —u sin v, 0). 

Here x uu = is obvious, since the w-parameter curves are straight lines. 
The v-parameter curves are helices, and this formula for the acceleration 
x vv was found already in Chapter II. Now by Lemma 4.2, 

„ (, x w X Xd / ^ 

t = x 



w 




\ x u X x„ ) 


b 


W 


W 


\ x u X x„ ) 


n 



Hence by Corollary 4.1 and the results above, 

L - m 2 -(b/W) 2 -b 2 -b 2 



K = 



EG - F 2 W 2 W i (b 2 + u 2 ) 2 



= Gl+ E n - 2F m = 

2(EG - F 2 ) 

Thus the helicoid is a minimal surface with Gaussian curvature 

-1 ^ K < 0. 

The minimum value K = — 1 occurs on the central axis (u = 0) of the 
helicoid, and K — > as distance | u \ from the axis increases to infinity. 

(2) The saddle surface M: z = xy (Example 1.3). This time we use 
the Monge patch x(u, v) = (u, v, uv) and with the same format as above, 
compute 



Sec. 4] 



x M = (1, 0, v) 
x v = (0, 1, u) 



COMPUTATIONAL TECHNIQUES 

E = 1 + v 2 

F = uv 



215 



U = (-v, -u,l)/W 
x uv = (0,0, 1) 



G = 1 + M 2 
TF = Vl + W 2 + V 2 

£= 
m = 1/W 
„ = 0. 



Hence 



# = 



-1 



(1 + u 2 + ?; 2 ) 2 ' 



# = 



— uv 



(1 + M 2 + V 2 ) 



3/2 



Strictly speaking, these functions are K{x) and #(x) denned on the 
domain E 2 of x. But it is easy to express K and H directly as functions on M 
by using the cylindrical coordinate functions r = \/x 2 + y 2 and z. Note 
from Fig. 5.27 that 

r{x{u,v)) = -\/u 2 + v 2 



and 



hence on M : 



z(x(u, v)) = uv; 



K = 



-1 



H = 



— z 



(1 + r 2 ) 2 ' (1 +r 2 ) 3 ' 2 

Thus the Gaussian curvature of M depends only on distance to the z 



P = x(u, v) 




FIG. 5.27 



216 SHAPE OPERATORS [Chap. V 

axis, rising from K = — 1 (at the origin) toward zero as r goes to infinity, 
while H varies more radically. 

Like all simple (that is, one-patch) surfaces, the helicoid and saddle 
surface are orientable, since computations as above provide a unit normal 
on the whole surface. Thus the principal curvature functions ki and k% are 
defined unambiguously on each surface. These can always be found from 
K and H by Corollary 3.5. Since the helicoid is a minimal surface, we get 
the simple result 



fci, h, = 



For the saddle surface, 

ki, h> = 



(6 2 + w 2 ) 

-z db Vl + r 2 + z 2 
(1 + r 2 ) 3 ' 2 



Techniques for computing principal vectors are left to the exercises. 

There is a different computational approach which depends on having 
an explicit formula Z = ^ ZiUi for a non vanishing normal vector field Z 
on M . The main case is a surface given in the form M : g = c, for there we 
know from Chapter IV, Section 3, that the gradient 

dXi 

is such a vector field — thus we may use any convenient scalar multiple of 
Vgf as Z. Let S be the shape operator derived from the unit normal 

u = z/\\ z ||. 

If V is a tangent vector field on M, then by Method 2 in Section 1, we find 

VyZ = X) V[Zi]Ui. 

Using a Leibnizian property of such derivatives, 



V V C7 = V, 



Z (VyZ) 



+ F [m] ; 



(Fig. 5.28). What is important here is that V[l/\\ Z \\]Z is a normal vector 
field; we do not care which one it is, so we denote it merely by — N v . Thus 

5(F)- -v vU = =l£p- + Ny. 

Note that if W is another tangent vector field on M, then N v X N w = 0, 
while products such as N v X Y are tangent to M for any Euclidean vector 



Sec. 4] COMPUTATIONAL TECHNIQUES 217 



Tangent plana 




V[l/\\ Z \\]Z 

FIG. 5.28 



field Y on M. Thus it is a routine matter to deduce the following lemma from 
Lemma 3.4. 

4.4 Lemma Let Z be a nonvanishing normal vector field on M . If V and 
W are tangent-vector fields such that V X W = Z, then 

R = (Z-VyZ X V W Z) 



H = -Z> 



II Z || 4 

(v v z x w + V X V W Z) 
211 Z II 3 



To compute, say, the Gaussian curvature of a surface M: g = c using 
patches, one must begin by explicitly finding enough of them to cover all of 
M; a complete computation of K may thus be tedious, even when g is a 
rather simple function. The following example shows to advantage the 
approach just described. 

4.5 Example Curvature of the ellipsoid 

2 2 2 

M: 9 = - 2 + | + - 2 = 1. 
a 2 o 2 c 2 

We write g = J2 Xi/ai, and use the (nonvanishing) normal vector field 

Z = ^g=j:^U i . 
at 

Now if V = J] «iC7, is a tangent vector field on M, 

«i a» 2 

since 



218 



SHAPE OPERATORS 



[Chap. V 



V[xi] = dxt(V) = vt. 
Similar results for another tangent vector field W then yield 



z*v v z x v w z = 



Xi 


X 2 
«2 2 


x 3 
a 3 2 


Vi 

«i 2 


V2 

a 2 2 


Vz 
a 3 2 




a 2 2 


w 3 
a 3 2 



1 



ai 2 a 2 2 a 3 2 



X-V X w. 



where X is the special vector field ^ XiUi which was used in Example 3.9 
in Chapter IV. 

It is always possible to choose V and W so that V X W = Z. 
But then 

2 

X*V X W = Z-Z = £— = !• 

a* 2 



Thus by Lemma 4.4 we have found 
1 



K = 



afatW II Z || 4 



where 



"(^7 



For an?/ oriented surface in E , its support function h assigns to each 
point p the orthogonal distance h(p) = pȣ/(p) from the origin to the 
tangent plane T P (M), as shown in Fig. 5.29 for the ellipsoid. Using the 
vector field X (whose value at p is the tangent vector p p ), we find for the 
ellipsoid that 

<7 1 

h = X'U = X-, 



z 




FIG. 5.29 



Sec. 4] COMPUTATIONAL TECHNIQUES 219 

Thus a more intuitive expression for the Gaussian curvature of the ellipsoid 
is 

K= h * 



a 2 b 2 c 2 



Note that when a = b = c = r, the ellipsoid is a sphere and this formula 
becomes K = 1/r 2 . 

The computational results in this section, though stated for surfaces, 
still apply to immersed surfaces (Exercise 10 of Chapter IV, Section 8). 
In particular, the formulas in Corollary 4.1 make sense for an arbitrary 
regular mapping x: D — > E 3 . The theoretical justification of this added 
generality is outlined in Chapter VII, Section 7. 



EXERCISES 

1. Show that the sphere of radius r has K = 1/r 2 by applying the methods 
of this section to the geographical patch 

x(u, v) = (r cos v cos u, r cos v sin u, r sin v). 

2. For a Monge patch, x(u,v) = (u, v,f(u,v)), show that 

E = 1 + fj t = f uu /W 

F = fufv m = fuv/W 

G = 1 + jv n = U/w 

where 

W = (1 +/„*■+ /."J 1 *. 
Find formulas for K and H. 

3. (Continuation). Deduce that the image of x is flat if and only if 

Juujvv Juv == "J 

minimal if and only if 

(1 + fu)f vv + (1 + /„ 2 ) /„„ - 2f u f v f uv = 0. 

4. Show that the image of the patch 

x(w, v) = (u, v, log cos v — log cos u) 
is a minimal surface with Gaussian curvature 



220 SHAPE OPERATORS [Chap. V 

„ _ —sec u secv 

where 

W 2 = 1 + tan 2 w + tanV 

5. Express the curvature K of the monkey saddle M: z = x 3 — 3xy 2 
(Fig. 4.47) in terms of r = s/x 2 + y 2 . Is this surface minimal? 

6. Find the Gaussian curvature of the elliptic and hyperbolic paraboloid, 

M:z = -2 + 8 C> < e = ±D- 

7. Show that the curve segment 

a(t) = x(d(0, <h(t)), a ^ t g b 
has length 

L(a) = f (Eai 2 + 2Fai a 2 ' + Ga^) m dt, 

•'a 

where 2?, i^, and (? are evaluated on 0,1,02. 

8. Prove that the coordinate angle # of a patch x: D — > M, < # < ir, 
is a differ entiahle function on D. (Hint: Use the Schwarz inequality in 
III) 

9. (a) A patch x in M is orthogonal provided F = (so x u and x„ are 

orthogonal at each point ) . Show that in this case 

&\Xu) == "^ X M -f- -^ X v 
0\X V ) = ^ X u -p — X„ . 

(b) A patch x is principal provided F = m = 0. Prove that x„ and x„ 
are principal vectors at each point, with corresponding principal 
curvatures l/E and n/G. 

10. Prove that a tangent vector 

V = ViX u + v 2 x v 

is a principal vector if and only if 

11 2 

= 0. 



2 

V2 


— Viv 2 


2 


jE7 


F 


G 


f 


m 


n 



(Hint: v is principal if and only if the normal vector aS(v) X v is zero.) 



Sec. 4] COMPUTATIONAL TECHNIQUES 221 

11. Show that on the saddle surface M (4.3) the two vector fields 

(Vi + u\ ± Vi + v 2 , i>Vi + u 2 ± wVi + y 2 ) 

are principal at each point. Check that they are orthogonal and tangent 
toM. 

12. (Enneper's minimal surface). This is the immersed surface given by 

t \ ( U . 2 *> 3 . 2 2 2\ 

x(w, v) = iu — — + UV , V — — + uv,u — V \. 

Prove that this immersed surface is minimal and that x is not one-to- 
one. {Hint: For H = it suffices to prove 

E = G, F = 0, and x uu + x„„ = 0.) 

13. (Patch criterion for umbilics). Show that the point x(u, v) is umbilic 
if and only if there is a number k such that t = kE, m = kF, and 
«. = kG at (u, v) (A; is then the principal curvature ki = k%). 

14. If v = vix u + V2X„ is tangent to M at x(w, v), the normal curvature in 
the direction determined by v is 

jf( \ _ ^ + 2mViV 2 + ««2 2 
~ j&yi 2 + 2FviV 2 + Gy 2 2 

where the various functions are evaluated at (u,v). 

15. Find the umbilic points (if any) on the following surfaces: 

(a) Saddle (Example 4.3). 

(b) Monkey saddle (Exercise 5). 

(c) Elliptic paraboloid (Exercise 6). 

16. (Tubes). If /? is a unit-speed curve in E 3 with k > 0, let 

x(u,v) = j3(u) + e(cos v N(u) + sin vB(u)). 

Thus the t;-parameter curves are circles of (constant) radius e in planes 
orthogonal to /?. Show that 

(a) x is regular if e is small enough; so x is an immersed surface called 
the tube' of radius e around /3. 

(b) U = cos v N(u) + sin v B(u) is a unit normal function on the 
tube. 

/ \ V —k(u) cos V 

\c) J\. = 



e(l — k(u) e cos v) 
(Hint: Use S(x u ) X S(x v ) = Kx u X x„.) 

17. Show that the elliptic hyperboloids of one and two sheets (Ex. 10 of 
IV.2) have Gaussian curvature K = —h*/a 2 b 2 c 2 and K = h /a 2 b 2 c 2 , re- 



222 



SHAPE OPERATORS 



[Chap. V 



spectively, and that both support functions h are given by the same 
formula as for the ellipsoid (4.5). 

18. If h is the support function of an oriented surface M (Z E 3 , show that 

(a) A point p of M is a critical point of h if and only if p-S(x) — 
for all tangent vectors v to M at p. (Hint: Write h as X* U, where 
X = J^XiUi.) 

(b) When K(p) ^ 0, p is a critical point of h if and only if p (con- 
sidered as a vector) is orthogonal to M at p. 

19. Use the preceding exercises to find the Gaussian curvature intervals 
of the ellipsoid and the elliptic hyperboloids of one and two sheets. 
(Ex. 10 of IV. 2) Assume a ^ b ^ c. 

20. Compute K and H for the saddle surface (Example 4.3) by the method 
given at the end of this section. (Hint: Take V and W tangent to the 
two sets of rulings of M. ) 

21. Scherk's minimal surface, M: e cos x = cos y. Let (R be the region 
in the xy plane on which cos x cos y > 0; (R is a checkerboard pattern 
of open squares, with vertices ((x/2) + irm, Or/ 2 ) + nir )- Show that 

(a) M is a surface. 

(b) For each point (u,v) in (R there is exactly one point (u, v, w) in M . 
The only other points of M are entire vertical lines over each of 
the vertices of (R (Fig. 5.30). 

(c) M is a minimal surface with K = -e z /(e z mxx + l) 2 . (Hint: 




FIG. 5.30 



Sec. 5] SPECIAL CURVES IN A SURFACE 223 

V = cos xUi + sin z£/ 3 is a tangent vector field.) Further proper- 
ties of this surface are given in Ex. 1 1 of VI.8. 

22. Let Z be a never-zero normal vector field on M. Show that a tangent 
vector v to M at p is principal if and only if 

vZ(p) XV,Z = 

(see the hint for Exercise 10). 
The equation above together with the tangency condition 

Z(p)-v = 

may be solved for the principal directions. Thus umbilics may be 
located using these equations: p is umbilic if and only if every tangent 
vector v at p is principal. 

23. Consider the ellipsoid M : ]£ x?/a? = 1. Show that 

(a) A tangent vector v at the point p is principal if and only if 

= piV2V S ((i2 2 — (h) + P2ViV 3 (a 3 2 — a?) + PzV\Vi(ai — a*). 



(b) Assuming a x > a 2 > a 3 , show that there are exactly four umbilics 
on M, with coordinates 

2\ 1/2 / 2 2\ 1/2 

# tti — a 

pi = ±ai 



(2 2\ 1/2 / 2 „ 2\ 1 

ai — a 2 \ n . /a2 — a 3 \ 



5 Special Curves in a Surface 

We shall briefly consider three geometrically significant types of curves in a 
surface M C E 3 . Neither this section nor the next is really essential for the 
theory; their purpose is to illustrate some of the ideas already introduced, 
and supply examples for later work. 

5.1 Definition A regular curve a in M C E 3 is a principal curve (or line 
of curvature) provided that the velocity a of a always points in a principal 
direction. 

Thus principal curves always travel in directions in which the bending 
of M in E 3 takes its extreme values. Neglecting changes of parametrization, 
there are exactly two principal curves through each nonumbilic point of 
M— and these necessarily cut orthogonally across each other. (At an umbilic 
point p, every direction is principal and near p the pattern of the principal 
curves can become quite complicated.) 

5.2 Lemma Let a be a regular curve in M CI E 3 , and let U be a unit 
normal vector field restricted to a. Then 



224 



SHAPE OPERATORS 



[Chap. V 



(1 ) The curve a is principal if and only if U and a are collinear at each 
point. 

(2) If a is a principal curve, then the principal curvature of M in the 
direction of a is a" • U/a*a. 

Proof. (1) Exercise 1 from Section 1 shows that S(a) = — U'. Thus 
U and a are collinear if and only if S(a ) and a are collinear. But by 
Theorem 2.5, this amounts to saying that a always points in a principal 
direction, or, equivalently, that a is a principal curve. 

(2) Since a is a principal curve, the vector field a /|| a || consists entirely 
of (unit) principal vectors belong to, say, the principal curvature ki. 
Thus 

h = *(«7|| «' ||) =S(a'/\\a'\\)-a'/\\a'\\. 

= S(a')-a = a"-U 
a! *a' a' *a' 

where the last equality uses Lemma 2.1. | 

In this lemma, (1 ) is a simple criterion for a curve to be principal, while 
(2 ) gives the principal curvature along a curve known to be principal. 

5.3 Lemma Let a be a curve cut from a surface M C E by a plane P. 
If the angle between M and P is constant along a, then a is a principal 
curve of M. 

Proof. Let U and V be unit normal vector fields to M and P (respectively) 
along the curve a, as shown in Fig. 5.31. Since P is a plane, V is parallel, 
that is, V = 0. Now the constant angle assumption means that U»V is 
constant; thus 

= (U'V)' = U''V. 

Since U is a unit vector, U' is orthogonal to U as well as V. The same is of 




FIG. 5.31 




Sec. 5] SPECIAL CURVES IN A SURFACE 225 

course true of a, since a lies in both M and P. If 
U and F are linearly independent (as in Fig. 
5.31) we conclude that U' and a are collinear; 
hence by Lemma 5.2, a is principal. 

However, linear independence fails only when 
U = ±F. But then if = 0, so a is (trivially) 
principal in this case as well. | 

(It is scarcely any harder to prove the gener- 
alization given in Exercise 5.) Using this result it FIG - 5 - 32 
is easy to see that the meridians and parallels of a 

surface of revolution M are its principal curves. Indeed, each meridian ju is 
sliced from M by a plane through the axis of revolution and hence orthog- 
onal to M along n, while each parallel ir is sliced from M by a plane orthog- 
onal to the axis, and by rotational symmetry such a plane makes a con- 
stant angle with M along ir. 

Directions tangent to M C E 3 in which the normal curvature is zero are 
called asymptotic directions. Thus a tangent vector v is asymptotic provided 
fc(v) = /S(v)»v = 0, so in an asymptotic direction M is (instantaneously, at 
least) not bending away from its tangent plane. 

Using Corollary 2.6 we can get a complete analysis of asymptotic direc- 
tions in terms of Gaussian curvature. 

5.4 Lemma Let p be a point of M C E . 

(1) If K(p) > 0, there are no asymptotic directions at p. 

(2) If K(p) < 0, then there are exactly two asymptotic directions at p 
which are bisected by the principal directions (Fig. 5.32) at angle & such 
that 

tan 2 * = :=ME- } . 
fe(p) 

(3) If K(p) = 0, then every direction is asymptotic if p is a planar 
point; otherwise there is exactly one asymptotic direction which is also 
principal. 

Proof. These cases all derive from Euler's formula 
A;(u) = h(p) cos 2 # + fc 2 (p) sinV 

in Corollary 2.6. 

(1) Since fci(p) and A; 2 (p) have the same sign, fc(u) is never zero. 

(2) Here fci(p) and /^(p) have opposite signs and we can solve the 
equation = fci(p) cosV + A- 2 (p) sin 2 ?? to obtain the two asymptotic 
directions. 

(3 ) If p is planar, then 



226 




[Chap. V 



FIG. 5.33 



h(p) = k»(p) = 0; 



hence Zs(u) is identically zero. If just 



then 



fc 2 (p) = 0, 



/c(u) = fci(p) cos 2 # 



is zero only when cos & = 0, that is, in the principal direction u = e 2 . | 

We can get an approximate idea of the asymptotic directions at a point 
p of a given surface M by picturing the intersection of the tangent plane 
f p (M) with M near p. When K (p) is negative, this intersection will consist 
of two curves through p whose tangent lines (at p ) are asymptotic direc- 
tions (Exercise 5 of Section 3 ) . 

Figure 5.33 shows the two asymptotic directions A and A at a point p 
on the inner equator of a torus. (The two intersection curves merge into a 
single figure-8.) 

5.5 Definition A regular curve a in M <Z E 3 is an asymptotic curve pro- 
vided its velocity a always points in an asymptotic direction. 

Thus a is asymptotic if and only if 

k(a) = S(a)-a = 0. 

Since S(a) = — U', this gives a criterion, U' »a = 0, for a to be asymp- 
totic. Asymptotic curves are more sensitive to Gaussian curvature than are 
principal curves: Lemma 5.3 shows that there are none in regions where K 
is positive, but two cross (at an angle depending on K) at each point of a 
region where K is negative. 

The simplest criterion for a curve in M to be asymptotic is that its ac- 
celeration a" always be tangent to M. In fact, differentiation of U*a =0 
gives 



Sec. 5] SPECIAL CURVES IN A SURFACE 227 

U'-a + U-a" = 0, 

so U'»a — (a asymptotic) if and only if U*a" = 0. 

The analysis of asymptotic directions in Lemma 5.4 has consequences 
for both flat and minimal surface. First, a surface M in E is minimal if 
and only if there exist two orthogonal asymptotic directions at each of its 
points. In fact, H (p ) = is equivalent to fci (p ) = — k 2 (p ) , and an examina- 
tion of the possibilities in Lemma 5.4 shows that /bi(p) = — A^(p) if and 
only if either (a) p is planar (so the criterion holds trivially) or (b) 

K(p) < withtf = ±x/4, 

which means that the two asymptotic directions are orthogonal. 

Thus a surface is minimal if and only if through each point there are two 
asymptotic curves which cross orthogonally. This observation gives geomet- 
ric meaning to the calculations in Example 4.3 which show that the helicoid 
is a minimal surface. In fact, the u- and v-parameter curves of the patch x 
are orthogonal since F = 0, and their accelerations are tangent to the surface 
since £ = U*x uu = and n = U*x vv = 0. 

Roughly speaking, a ruled surface M is one which is swept out by a straight 
line moving through E — the various positions of the line are called the 
rulings of M. Thus M has a parametrization in the ruled form 

x(w, v) = j8(w) + v8(u), or /3(v) + u8(v) 

where /3 and 5 are curves in E with 5 never (see Exercise 4-9 of Section 2 
of Chapter IV). For example, the helicoid is a ruled surface, since the 
patch in Example 4.3 may be written as 

x(w, v) = (0, 0, bv) + w(cos v, sin v, 0). 

This shows how the helicoid is swept out by a line rotating as it rises along 
the z axis. The saddle surface M : z = xy is doubly ruled, since 

, x / x f(u, 0, 0) + v(0, l,u) 

It is no accident that both these surfaces have K negative, for: 

5.6 Lemma A ruled surface M has Gaussian curvature K ^ 0. Further- 
more K = if and only if the unit normal U is parallel along each ruling of 
M (so all points p on a ruling have the same tangent plane f p (M).) 

Proof. A straight line t —> p + tq in any surface is certainly asymptotic, 
since its acceleration is zero, and thus trivially tangent to M . By definition, 
a ruled surface contains a straight line through each of its points. Hence 
there is an asymptotic direction at each point, so by Lemma 5.4, K fS 0. 



228 SHAPE OPERATORS [Chap. V 

Now let a (t) = p + tq be an arbitrary ruling in M. If U is parallel along 
a, then S (a ) = U = 0. Thus a is a principal curve with principal curva- 
ture k(a) = 0, and so K = k^ = 0. 

Conversely, if K = 0, we deduce from the case (3) in Lemma 5.4 that 
asymptotic directions (and curves) in M are also principal. Thus each rul- 
ing a is principal (S(a) = k(a)a) as well as asymptotic (k(a) = 0): 
hence U' = -S(a) = 0. | 

We come now to the last and most important of the three types of curves 
under discussion. 

5.7 Definition A curve a in M <Z E 3 is a geodesic of M provided its ac- 
celeration a" is always normal to M. 

Since a" is normal to M , the inhabitants of M perceive no acceleration at 
all — for them the geodesic a is a "straight line." A full study of geodesies 
is given in Chapter VII, where in particular we examine their character as 
shortest routes of travel. Geodesies are far more plentiful on a surface M 
than are principal or asymptotic curves; indeed by Theorem 4.2 of Chapter 
VII there is a geodesic through every point of M in every direction. 

Since its acceleration a" is, in particular, orthogonal to its velocity a, a 
geodesic a has constant speed, for differentiation of 

|| a || 2 = a *a yields 2a •«" = 0. 

A straight line a(t) = p + U\ in M is always a geodesic of M, since its 
acceleration a" = is trivially normal to M. Unlike principal and asymp- 
totic curves, geodesies cannot be defined in terms of the shape operator; 
however, a (unit speed) geodesic a with positive curvature bears an in- 
teresting relation to S, which uses the Frenet apparatus of a. Since the 
principal normal N = a" '/k of a is normal to the surface M, we have 

-N' = S(a) = S(T). 

Thus by a Frenet equation, S(T) = kT - tB. 

These remarks are sufficient to derive the geodesies of three rather special 
surfaces. 

5.8 Example Geodesies on some surfaces in E . 

(1) Planes. If a is a geodesic in a plane P orthogonal to u, a «u = 0, 
hence a" *u = 0. But a" is by definition normal to P; hence a." = 0. Thus 
a is a straight line, and since every such line is a geodesic, we conclude 
that the geodesies of P are all straight lines in P. 

(2) Spheres. If a is a (unit speed) geodesic in a sphere S of radius r, 
then, by a remark above, S(T) = kT — tB. (We saw in Chapter II, Sec- 
tion 3, that any curve in S has positive curvature, so the Frenet apparatus 



Sec. 5] 



SPECIAL CURVES IN A SURFACE 



229 




FIG. 5.34 

is available.) But Example 1.3 shows that S(T) = ±(\/r)T, depending 
on which of the two unit normals is used. These two equations for S(T) 
imply that k = \/r and t = 0. Hence by Lemma 3.6 of Chapter II, a lies 
on a circle C of radius r. This maximum radius r forces C to be a great circle 
of 2, that is, one sliced from 2 by a plane through its center. Conversely, 
any constant-speed curve running along a great circle has its acceleration 
a" point toward the center of that circle, which is also the center of 2, so 
a." is normal to 2. We conclude that the geodesies of 2 are the constant-speed 
parametrizations of its great circles (Fig. 5.34). 

(3) Cylinders. The geodesies of, say, the circular cylinder M:x 2 -\- y 2 = r 2 
are all curves of the form 

a(t) = (r cos (at + b), r sin (at + b), ct + d). 

In fact, any curve in M may be written 

a(t) = (rcos t?(0, r sin d(t), h(t)), 

and a vector normal to M has z-coordinate zero. Thus if a is a geodesic, 
then h" = 0, so h(t) = ct + d. The speed (rV 2 + fi*) 1 ' 2 of a is constant, 
so & is constant; hence &(t) = at + 6. 

When the constants a and c are both nonzero, a is a helix on M. The 
extreme case a = gives the rulings of M, and c = gives the cross- 
sectional circles. 

The essential properties of the three types of curves we have considered 
may be summarized as follows : 

Principal curves k (a ) = ki or k 2 S (a ) collinear a 

Asymptotic curves k(a) = S (a) orthogonal a a" tangent M 

Geodesies a" normal M 



EXERCISES 

1. Prove that a curve a in M is a straight line of E 3 if and only if a is 
both geodesic and asymptotic. 



230 



SHAPE OPERATORS 



[Chap. V 




2. To which of the three types — principal, asymp- 
totic, geodesic — do the following curves belong? 

(a) The top circle a on the torus (Fig. 5.35). 

(b) The outer equator /3 of the torus. 

(c) The x axis on M : z = xy. 
(Assume a constant-speed parametrization. ) 

3. On a surface of revolution, show that all meridians are geodesies, but 
that the parallel through a point a(t) of the profile curve is a geodesic 
if and only if a (t) is parallel to the axis of revolution. 

4. Let a be an asymptotic curve in M <Z E 3 with curvature k > 0. 

(a) Prove that the binormal B of a is normal to the surface along a, 
and deduce that S(T) = tN. 

(b) Show that along a the surface has Gaussian curvature K = — r 2 . 

(c) Use (b) to find the Gaussian curvature of the helicoid (Example 
4.3). 

5. Suppose that a curve a lies in two surfaces M and M which make a 
constant angle along a ([/• £7 constant). Show that a is principal in M 
if and only if principal in M. 

6. If x is a patch in M, prove that a curve a(t) = x(ai(t), cte(0) is 
(a) Principal if and only if 



E 



/2 



— ai ai a\ 
F G 



= 0, 



(b) Asymptotic if and only if fa/ 2 + 2mai'a 2 / + n.a^ = 0. 

Let a be a unit-speed curve inlc E 3 . Instead of the Frenet frame 

field on a, consider the frame field T, V, , U — where T is the unit tangent 

of a, U is the surface normal restricted to a, and V = U X T (Fig. 

5.36). 

(a) Show that 

T' = gV + kU U 

V' = -gT + tU 

U' = -kT - tV. 

where k = S(T)*T is the nor- 
mal curvature k(T) of M in the 
T direction, and t = S(T)'V. 




FIG. 5.36 



Sec. 5] SPECIAL CURVES IN A SURFACE 231 

The new function g is called the geodesic curvature of a. 
(b) Deduce that a is 

geodesic <=» g = 

asymptotic <=> fc = 

principal <=> tf = 0. 

8. If a is a (unit speed) curve in M, show that 

(a) a is both principal and geodesic if and only if it lies in a plane 
everywhere orthogonal to M along a. 

(b) a is both principal and asymptotic if and only if it lies in a plane 
everywhere tangent to M along a. 

9. On the monkey saddle M (Ex. 5 of IV.4) find three asymptotic curves 
and three principal curves passing through the origin 0. (This is possi- 
ble because only because is a planar umbilic point.) 

10. Show that the ruled surface x(u, v) = /3(w) + v8(u) has Gaussian 
curvature 

K== -m 2 -(js'-a x b'y 



EG - F 2 W 4 



where 



W =-||/?' X 8 + v8' X«||. 
11. (Flat ruled surfaces). 

(a) Show that cones and cylinders are flat (see Exs. 5 and 6 of IV.2) 

(b) If is a unit-speed curve in E 3 with k > 0, the ruled surface 

x(u,v) = 0(w) + vT(u), v > 0, 

is called the tangent surface of /? (Fig. 5.37). Prove that x is regu- 
lar, and that the tangent surface is flat. 




FIG. 5.37 



232 SHAPE OPERATORS [Chap. V 

12. Let a be a regular curve in M C E 3 , and let U be the unit normal of 
M along a. Show that a is a principal curve of Af if and only if the ruled 
surface x(u, v) = a(u) + vU(u) is flat. 

13. A closed geodesic of M is a geodesic which is a periodic function 
a : R — * M . Find all closed geodesies in a sphere, a plane, and a circular 
cylinder. 

14. A ruled surface is noncylindrical if its rulings are always changing 
directions; thus for any director curve, 8 X 8' ^ 0. Show that 

(a) a noncylindrical ruled surface has a parametrization 

x(u, v) = a(u) + v8(u) 

for which || 5 || = 1 and a •$' = 0. 

(b) for this parametrization, 

v -p\u) <x'-b X 8' 

& = / o/ x — ; — ^r« where p = ; — ; — . 

(p\u) + v 2 ) 2 F 8 f »8' 

The curve a is called the striction curve, and the function p is the 
distribution parameter. 

(Hint: For (a), if || 5 || = 1, find /so that a = a + f8. For (b), show 
a X 5 = pS'.) 

15. Describe the qualitative behavior of Gaussian curvature K on an 
arbitrary ruling of a (noncylindrical) ruled surface. Show that the 
route of the striction curve is independent of the choice of parametriza- 
tion, and that the distribution parameter is essentially a function on 
the set of rulings. 

16. Show that the striction curve of the helicoid is its central axis, and that 
its distribution function is constant. 

1 7. Find the striction curve and distribution parameter for 

(a) Both sets of ruling of the saddle surface (Example 4.3) 

(b ) Both sets of rulings of the hyperboloid of revolution 

2,2 2 

M: — = 1 

a 2 b 2 

(Fig. 5.38) (Find a ruled parametrizetion by modifying 
Ex. 9 of IV.2.) 

18. If x(u,v) = a(u) + v8(u) parametrizes a noncylindri- 
cal ruled surface, let L(u) be the ruling through a(u). 
Show that 

(a) If t? £ is the (smallest positive) angle from L(u) to 
L(u + e), and 




FIG. 5.38 



Sec 5] 



SPECIAL CURVES IN A SURFACE 



233 



d e is the orthogonal distance from L(u) to L{u + e), then lim (d E / 

£-»0 

# e ) = p(w). Thus the distribution parameter is the reciprocal of 
the rate of turning of L — its sign describes the direction of the 
turning.) 

(b) There is a unique point p E of L(u) which is nearest to L(u + e), 
and lim p e = a(u). (This gives another characterization of the 

£-»0 

striction curve.) 

(c) The distance from a(u) to a(u + e) need not be a good approxi- 
mation of the distance d e from L(u) to L(u + e). Give an example. 

19. Let x(u,v) = a + v8, || 5 || = 1, parametrize & flat ruled surface M. 
Show that 

(a) If a' is always zero, then M is a cone.f 

(b) If h' is always zero, then M is a cylinder. 

(c) If both a and 5' are never zero, then M is the tangent surface of 
its striction curve. (Exercise lib.) 

Of course these cases are far from exhausting all the possibilities, 
but in a sense they show that an arbitrary flat ruled surface ("de- 
velopable surface") is a mixture of the three types in Exercise 11. If 
such a surface is closed in E 3 (Ex. 10 of IV.8), it must be a cylinder, 
since the closure condition implies that the rulings are entire straight 
lines. 

20. A right conoid is a ruled surface whose rulings all pass through a fixed 
axis (Fig. 5.39). Taking this axis as the z axis of E 3 , we get the para- 
metrization 

x(u,v) = (u cos $(v), u sin &(v), h(v)) 

(a) Find the Gaussian and mean curvature. 

(b) Show that the surface is noncylindrical if 
i?' is never zero; find its striction curve 
and parameter of distribution. 

21. Sketch the conoid M parametrized by 

x(u, v) = (u cos v, u sin v, cos 2 v), 

and find its Gaussian and mean curvature. 
Express M in the form z = f(x,y) (origin 
omitted). 

22. Prove that a surface which is both ruled and fig. 5.39 
minimal is part of either a plane or a helicoid. 

t In each case, we assume there are no restriction on v except those necessary 
to ensure that x is regular. 




234 SHAPE OPERATORS [Chap. V 

{Hint: Flat regions in M are planar; thus arguing as in Theorem 6.2 
we may suppose K < 0. Use the parametrization in Exercise 14 with 
the additional feature that 8 is a unit-speed curve. Then H = leads 
to three equations. Deduce that 8 is a unit circle; we may assume 
8(u) = (cos u, sin u, 0).) 



6 Surfaces of Revolution 

The geometry of a surface of revolution is rather simple, yet these surfaces 
exhibit a wide variety of geometric behavior; thus they offer a good field 
for experiment. 

We shall apply the methods of Section 4 to study an arbitrary surface of 
revolution M, parametrized as in Example 2.5 of Chapter IV by 

x(u,v) = (g(u), h(u) cos v, h(u) sin v). 

Recall that h(u) is the radius of the parallel of M at distance g(u) along 
the axis of revolution, as shown in Fig. 4.17. This geometric significance of 
g and h means that our results do not depend on the particular position of 
M relative to the coordinate axes of E 3 . 

Because g and h are functions of u alone, we write 

x u = (</', h' cos v, h' sin v) E = g' 2 + h' 2 

F = 
x„ = (0, —h sin v, h cos v) G = h 2 . 

Here E is the square of the speed of the profile curve — and of all meridians 
(w-parameter curves) — while G is the speed squared of the parallels (v- 
parameter curves). Next 

x M X x„ = (hh , — hg cos v, —hg sin v) 

|| x u X x v || = VEG^F 2 = hVg' 2 + h' 2 

U = (h', —g cos v, —g sin v)/vV 2 + h' 2 . 

Taking second derivatives, we obtain 

x uu = (g", h? cos v, h" sin v) 

x™ = (0, — h sin v, h cos v) 

Xw = (0, —h cos v, —h sin v) 

I = i-gh" + g"h')/Vg' 2 + h' 2 

m = 

n = gh/Vg' 2 + h' 2 



Sec. 6] 



SURFACES OF REVOLUTION 



235 



Since F — m = 0, it is easy to check (Exercise 9 of Section 4) that for 
the shape operator S derived from U, 



0\X.u) J-, X-i 



ld\X v J — ™ \ v . 



(Thus we have an analytical proof that the meridians and parallels of M 
are its principal curves.) Hence if the corresponding principal curvature 
functions are denoted by k„. and K (instead of k\ and fc 2 ), we have 

g' h' 



fCu — ^ — 



g" h" 



E (g' 2 + h' 2 ) 3 ' 2 ' 
Thus the Gaussian curvature of M is 



/C?r — "7S — 



9 



K - Mv - T(gf2 + h , 2)2 



G h(g' 2 + h' 2 ) 1 ! 2 



g' h' 

a" h" 



(1) 



(2) 



Note that this formula defines K as a real- valued function on the domain 
/ of the profile curve 

oc(u) = (g(u), h(u),0). 

By the conventions in Section 4, K(u) is the Gaussian curvature K(x(u,v)) 
of every point on the parallel through a(u). Similarly, with the other functions 
above — because of the rotational symmetry of M about the axis of revolu- 
tion, its geometry is "constant on parallels" and completely determined by 
the profile curve. 

In the special case when the profile curve passes at most once over each 
point of the axis, we can usually arrange for the function g to be simply 
g(u) = u (Exercise 13 of Chapter IV, Section 2). Thus the formulas 
above reduce to 



rCii — 



k r = 



K = 



-h" 



(1 + h' 2 ) 3 ' 2 
1 

h(i + h' 2 y 2 
-v 



(3) 



h(l + h' 2 ) 2 ' 
6.1 Example Surfaces of Revolution 

(1) Torus of revolution T. The parametrization x in Example 2.6 of 
Chapter IV has 



236 SHAPE OPERATORS [Chap. V 

g(u) = r sin u, h(u) = R + r cos u. 

Although the axis of revolution is now the z axis, formulas (1) and (2) 
above remain valid and we compute 

E = r 2 F = G = (R + r cos uf 
£ = r m = n = (R + r cos u) cos u 

*. = ! *,= cosw 



r i? + r cos w 

cos u 



K = 



r(R + r cos w) ' 

We now have an analytical proof that K is positive on the outer half 
of the torus and negative on the inner half. In fact K has its maximum 
value l/r(R + r) on the outer equator (u = 0) and its minimum value 

-l/r(R - r) 

on the inner equator (u = T ), and K is zero on the top and bottom circles 

(u = ±tt/2). 

(2) Catenoid. The curve y = c cosh (x/c) is a catenary; its shape is that 
of a chain hanging under the influence of gravity. The surface obtained by 
revolving this curve about the x axis is called a catenoid (Fig. 5.40). From 
the formulas in (3 ) we get 

/C u = K T = 



c cosh 2 (u/c) 
Hence 



K- - 1 



c 2 cosh 4 (w/ c ) 



Since its mean curvature H is zero, the catenoid is a minimal surface. 
Its Gaussian curvature interval is — 1/c 2 5j K < 0, with minimum value 
& = — 1/c on the circle w = 0. 

The following result shows that catenoids are the only complete] non- 
planar surfaces of revolution which are minimal. (A plane is trivially a 
minimal surface, since k\ = fc 2 = 0. ) 

6.2 Theorem If a surface of revolution M is a minimal surface, then M 
is contained in either a plane or a catenoid. 

t We use this word with its dictionary meaning; a mathematical definition is 
given in Section 4 of Chapter VII. 



Sec. 6] 



SURFACES OF REVOLUTION 



237 







FIG. 5.40 



Proof. We use the parametrization 

x(u, v) — (g(u), h(u) cos v, h(u) sin v) 

of M, with u in some interval /, and v arbitrary. 

Case 7. g is identically zero. Then g is constant, so M is part of a plane 
orthogonal to the axis of revolution. 

Cose 2. g is never zero. Then by an earlier exercise, M has a parametriza- 
tion of the form 

y(u, v) = (u,f(u) cos v,f(u) sin v). 

The formulas for k x and k 2 in (3) on page 235 then show that the minimal- 
ity condition ki + k 2 = is equivalent to 

//" = 1+/' 2 . 
Because u does not appear explicitly in this equation, there is a standard 
method for solving it, which may be found in any elementary text on 
differential equations. We merely record that the solution is 



f(u) = a cosh 



M- 



where a and b are arbitrary constants. Thus M is part of a catenoid. 

Cose 3. g is zero at some points, nonzero at others. This cannot occur. 
For definiteness suppose that g ' (u ) = 0, but g (u) > f or u < u . By 
case 2, the profile curve 

«(w) = (g(u), h(u),0) 



238 SHAPE OPERATORS [Chap. V 

is a catenary for u < u . But the shape of this curve makes it clear that its 
slope h /g cannot suddenly become infinite at u = uq. | 

Helicoids and catenoids are the "elementary" minimal surfaces. Two 
others are given in Exercises 12 and 21 of Section 4. Soap-film models of 
an immense variety of minimal surfaces may easily be constructed by the 
methods given in Courant and Robbins [4], where the term minimal is 
explained. 

The expression vV 2 + ^' 2 which appears so frequently in the general 
formulas at the beginning of this section is, of course, just the speed of the 
profile curve 

«(«) = (9(u), h(u),0). 

Thus we can radically simplify matters by replacing a by a unit-speed 
reparametrization. The resulting surface of revolution M is unchanged: 
We have merely given it a new parametrization, said to be canonical. 

6.3 Lemma If x is a canonical parametrization of a surface of revolution 

M (g 2 + ti 2 = 1), then 

E = 1, F = 0, G = h 2 
and 

-h" 



K = 



h 



Proof. Since g' 2 + h! 2 = 1, these expressions for E, F, and G follow im- 
mediately from those on page 234, and K in (2) becomes 



* = ^f 



h' 



_ -g hr + g g"h 



h 



Differentiation of g 2 = 1 — h' 2 yields g g" = —h h" ; hence 

„ -(1 - h ,2 )h" - h' 2 h" hT - 

K = r = 7- • I 

h h 

The effect of canonical parametrization is to shift the emphasis somewhat 
from measurements in the space outside M (for example, along the axis 
of revolution) to measurements within M itself. This idea will be developed 
more fully in Chapters VI and VII. 

6.4 Example Canonical parametrization of the catenoid (c = 1). 
An arc-length function for the catenary 

a(u) = (u, cosh u) is s(u) = sinh u; 



Sec. 6] SURFACES OF REVOLUTION 239 




}*(•) 



9(s) 
FIG. 5.41 

hence a unit-speed reparametrization is 

0( 8 ) = (g(s),h(s)) = (sinh _1 s, y/l + s 2 ) 

as indicated in Fig. 5.41. The resulting canonical parametrization of the 
catenoid is then 

x(s, t>) = (sinh _1 s, y/l + s 2 cos v, y/\ + s 2 sin v). 

Thus by the preceding lemma 

w x _ ^) _ "I 
AW ~ />(«) (1 +s 2 ) 2 " 

This formula for Gaussian curvature in terms of x is consistent with the 
formula K(u) = — l/cosh 4 w found in Example 6.1 for the parametrization x. 
In fact, since s(u) = sinh u, we have 

K(s(w)) = (1 +s 2 (w)) 2 = (1 +sinh 2 w) 2 = cosh 4 w ' 

The simple formula for K in Lemma 6.3 suggests a way to construct 
surfaces of revolution with prescribed Gaussian curvature. Given a func- 
tion K on some interval, we first solve the differential equation h" + Kh = 
for h, subject to initial conditions h(0) > and | h' (0) \ < 1. To get a 
canonical parametrization we need a function g such that g + h =1. 
Evidently 



g{u) = f Vl - h"(t) dt 



will do the job. Thus for any interval around on which the conditions 
h > and | h' | < 1 both hold, we can revolve 

a(u) = (g(u), h(u), 0) 

around the x axis to obtain a surface of revolution whose Gaussian curva- 
ture is, by Lemma 6.3, precisely —h"/h = K. 

6.5 Example Surfaces of revolution with constant positive curvature. 



240 SHAPE OPERATORS 



[Chap. V 



We apply the procedure above to the constant function K = 1/c 2 The 
differential equation h" + (l/c)h = has general solution 



h(u) = a cos (- + 6 J 



The constant b represents merely a translation of coordinates; we may 
suppose that b = and a > 0. Thus the functions 



gin) = f U a/ 1 _<L 



h(u) = a cos - 



sin - dt 
c 



give rise to a surface of revolution M a with constant Gaussian curvature 
K = 1/c . The necessary conditions h > and | ti \ < 1 determine the 
interval / to which u must be restricted. The constant c is fixed throughout 
the discussion, but the constant a is at our disposal; we consider three cases. 
Cose 1. a = c. Here 

g(u) = / cos - dt = c sin - , 
J o c c 

and h(u) = c cos (u/c). The interval i" is thus -ttc/2 < u < T c/2. Since 
this profile curve a(u) = (g(u), h(u)) is a semicircle, revolution about 
the x axis produces the sphere 2 of radius c— except for its two points on 
the x axis. 

Cose 2. < a < c (Fig. 5.42). Here h is positive on the same interval 
as above, and | ti \ < 1 is always true, so g is well-defined. The profile 
curve u -> (g(u), h(u)) still has the same length ttc/2, but it now makes 
a shallower arch, which rests on the x axis at ±a*, where 



4 C KV^!< 



Approximate values for this elliptic integral may be found in tables, 
but clearly as a shrinks from c down to 0, a* increases from c to xc/2. The 




Sec. 6] SURFACES OF REVOLUTION 241 

resulting surface of revolution M a — round for a = c — becomes football- 
shaped at first, and for a very small is a needle of length just less than xc/2. 
By contrast with Case 1, the intercepts (±a*, 0, 0) cannot be added to 
M a now, since this surface is actually pointed at each end. The differential 
equation h," + (1/c 2 ) A = has delicately adjusted the shape of M a so 
that its principal curvatures are no longer equal, but still give 

K = k li k r = 1/c 2 . " 



Case 3. a > c (Fig. 5.43). Here the inter- 
val / is reduced in size, since the expression 
under the square root (in the formula for g) /^ ,i \ 

becomes zero at £*such that sin (t /c) = c/a "" 

< 1. 

Thus 



Va 2 - (?/ 



($) - * 



h(t*) = a cos I — J = v a 2 — c 2 . fig. 5.43 



As a increases from a = c, the resulting surface of revolution M a is at first 
somewhat like the outer half of a torus, but when a is very large, it becomes 
an enormous circular band whose very short profile curve is sharply curved. 
(kp must be large, since K ~ I /a is small and k,Jk r = 1/c .) 

A similar analysis for constant negative curvature leads to an infinite 
family of surfaces of revolution with K = — 1/c 2 (Exercise 9). The simplest 
of these is 

6.6 Example The bugle surface B. The profile curve of B (in the xy 
plane) is characterized by this geometric condition: It starts at the point 
(0, c) and moves so that its tangent line always reaches the x axis after 
running for distance exactly c. This curve (a tractrix) may thus be de- 
scribed analytically by a(u) — (u, h(u)), u > 0, where h is the solution 
of the differential equation h = —h/y/c 2 — h 2 such that h(u) -^ c as 
u — > 0. (The resulting surface of revolution is shown in Fig. 5.44.) 

Using this differential equation, we deduce from the formulas (3) on 
page 235 that the principal curvatures of B are 

k -=!l k - 1 

c ch 

Thus the bugle surface (or tractroid) has constant negative curvature 

K = -1/c 2 . 

This surface cannot be extended across its rim (not part of B) to form a 
larger surface in E 3 , since A* M (w) — > °o as u — -> 0. 



242 



SHAPE OPERATORS 



[Chap. V 




FIG. 5.44 



When this surface was first discovered, it seemed to be the analogue, 
for K constant negative, of the sphere; it was thus called a pseudosphere. 
However as we shall see in Chapter VII the true analogue of the sphere is 
quite a different surface and cannot be found in E 3 . 



EXERCISES 

1. Find the Gaussian curvature of the surface obtained by revolving the 
curve y = e~ x /2 around the x axis. Sketch this surface and indicate the 
regions where K > and K < 0. 

2. (Sign of Gaussian curvature). When y = f(x) is revolved about the x 
axis, and K is expressed in terms of x, show that K has the same sign 
( — , 0, + ) as —/" for each value of x. So K is positive on parallels through 
convex intervals on the profile curve, and negative for concave intervals, 
as shown Fig. 5.45. (The same result is true for an arbitrary surface of 
revolution, with convexity and concavity taken relative to the axis of 
revolution A.) 

3. (Magnitude of Gaussian curvature). Show that 

I K( u ) I — curvature k(u) of profile curve a at a(u), 

| k r (u) | = h(u) | cos <p(u) |, where <p(u) is the slope angle of the 

profile curve at a(u), 
hence that 

| K | = ich | cos <p | . 



Sec. 6] 



SURFACES OF REVOLUTION 



243 



K > 



K > 




K < 



-4 

FIG. 5.45 



tf > 




K = 



X <0 




FIG. 5.46 

4. An elliptical torus M is obtained by revolving an ellipse 

(x - R) 2 /a + 2/V& 2 = 1 

about the y axis (R > a). Find a parametrization for M and compute 
its Gaussian curvature. (Check your answer by setting a = b = r.) 

5. If r = *\/x 2 + y 2 > is the usual polar-coordinate function on the xy 
plane, and / is a differentiate function, show that M : z = f(r) is a 
surface of revolution, and that its Gaussian curvature (expressed in 
terms of r) is 

K ^ f(r)f''(r) 

r(l +/'(r) 2 ) 2 " 

6. Find the Gaussian curvature of the surface M: z = e~ r /2 . Sketch this 
surface, indicating the regions where K > and K < 0. 

7. Prove that a flat surface of revolution is part of a cone or a cylinder. 

8. Let M be the surface obtained by revolving one arch ( — -k < t < ir) 
of the cycloid 7(<) = (t + sin t, 1 -f cos t) about the x axis. 

(a) Compute K relative to the usual parametrization of M. 

(b) Find the function h giving the height of a in terms of its arc length 

(measured from the top of the arch). Computed = —h"/h. 
{Hint: Use half -angles.) 

(c) Show that the results in (a) and (b) are consistent. 



244 SHAPE OPERATORS [Chap. V 

9. (Surfaces of revolution with constant negative curvature K = — 1/c 2 ). 
As in the case K = 1/c 2 , there is a family of such surfaces, divided into 
two subfamilies by a special surface. The solutions of 

h" - h/c 2 = 

giving, by canonical parametrization, essentially all these surfaces are: 

(a) h(u) = a sinh (u/c), < a < c, u > 0. Show that the profile curve 
a (u) = (g(u), h(u)) leaves the origi n with s lope a/\/ c 2 — a 2 and 
rises toward a maximum height of \/c 2 — a 2 . Sketch the resulting 
surface of revolution M a for a small value of the parameter a, and 
for a near c. 

(b) h(u) = b cosh (u/c), b > 0. Show that the profile curve a rises 
symmetrically (for ±w) toward a maximum height of s/c 2 + b 2 . 
Sketch the resulting surface M b for b small and for b large. 

(c) h(u) = ce u,c , u < 0. This surface B is merely a mirror image of 
the surface B gotten from 



h(u) = ce , u > 0. 

Show that B is, in fact, the bugle surface (Example 6.6). How does 
the surface B separate the two subfamilies, that is, for which values 
of a and b do M a and M b resemble B? (See Fig. 5.46 where M a and 
M b have been translated along the axis of revolution.) 



7 Summary 

The shape operator S of a surface M in E 3 measures the rate of change of 
a unit normal U in any direction on M. If we imagine U as the "first 
derivative" of M, then S is the "second derivative." But the shape opera- 
tor is also an algebraic object consisting of linear operators on the tangent 
planes of M . And it is by an algebraic analysis of S that we have been led 
to the main geometric invariants of a surface in E 3 : its principal curvatures 
and directions, and its Gaussian and mean curvatures. 



CHAPTER 



VI 



Geometry of Surfaces in E 3 



Now that we know how to measure the shape of a surface M in E , the 
next step is to see how the shape of M is related to its-other properties. Near 
each point of M, the Gaussian curvature has a strong influence on shape 
(Remark 3.3 of Chapter V), but we are now interested in the situation in 
the large — over the whole extent of M. For example, what can be said about 
the shape of M if it is compact, or flat, or both? 

Almost 150 years ago Gauss raised a question that led to a new and 
deeper understanding of what geometry is: How much of the geometry of 
a surface in E 3 is independent of its shape? At first glance this seems a strange 
question — what can we possibly say about a sphere, for example, if we 
ignore the fact that it is round? To get some grip on Gauss's question, let 
us imagine that the surface McE s has inhabitants who are unaware of 
the space outside their surface, and thus have no conception of its shape in 
E 3 . Nevertheless, they will still be able to measure "the distance from place 
to place in M and find the area of regions in M. In this chapter and the 
next we shall see that in fact they can construct an "intrinsic geometry" 
for M that is richer and no less interesting than the familiar Euclidean 
geometry of the plane E 2 . 



1 The Fundamental Equations 

To study the geometry of a surface M in E we shall apply the Cartan 
methods outlined in Chapter II. As with the Frenet theory of a curve in 
E 3 , this requires that we put frames on M, and examine their rates of change 
along M . Formally, a Euclidean frame field on M C E consists of three 

245 



246 GEOMETRY OF SURFACES IN E 3 [Chap. VI 

E, 




FIG. 6.1 

Euclidean vector fields (Definition 3.7, Chapter IV) that are orthonormal 
at each point. Such a frame field can be fitted to its surface as follows. 

1 .1 Definition An adapted frame field E h E 2 , E 3 on a region in M CZ E 3 
is a Euclidean frame field such that E 3 is always normal to M (hence E\ 
and E 2 are tangent to M) (Fig. 6.1). 

Thus the normal vector field denoted by U in the preceding chapter now 
becomes E 3 . For brevity we shall refer to an adapted frame field "on M," 
but the actual domain of definition is in general only some region in M, 
since an adapted frame field need not exist on all of M. 

1.2 Lemma There is an adapted frame field on a region in M a E 3 if 
and only if is orientable and there exists a nonvanishing tangent vector 
field on 0. 

Proof. This condition is certainly necessary, since E 3 orients 0, and E x 
and E 2 are unit tangent vector fields. To show that it is sufficient, let be 
oriented by a unit normal vector field U, and let V be a tangent vector 
field that does not vanish on 0. But then it is easy to see that 

Ex = Sj , E 2 = U XE h E 3 = U 

is an adapted frame field on 0. | 

1.3 Example Adapted frame fields. 

(1) Cylinder M: x 2 + y = r 2 . The gradient of g = x 2 + y 2 leads to the 
unit normal vector field E 3 = (xUi + yU 2 )/r. Obviously the unit vector 
field U% is tangent to M at each point. Setting E 2 = U 3 X E 3 , we then 
get the adapted frame field 

Ey = U 3 

-yVx + xU 2 
r 

xUi + yU 2 

Hj 3 = 

r 
on the whole cylinder M (Fig. 6.2). 



Sec. 1] 



THE FUNDAMENTAL EQUATIONS 



247 



(2) Sphere 2: x + y 2 + z = r\ The outward 
unit normal 



E% = 



xUx + yU 2 + zU* 



is defined on all of 2, but as we shall see in 
Chapter VII, every tangent vector field on 2 
must vanish somewhere. For example, the "due 
east" vector field V = — yUx + xU 2 is zero at the 
the north and south poles (0,0, ±r). Thus the 
adapted frame field 



E x = 



V 



M 




-E, 



K E t 



FIG. 6.2 



Ei = E% X E\ 



Es = 



xUx + yU 2 + zU 3 



(Fig. 6.3) is defined on the region in 2 gotten by deleting the north and 
south poles. 

Lemma 1.2 implies in particular that there is an adapted frame field on 
the image x(£>) of any patch in M; thus such fields exist locally on any 
surface in E . 

Now we shall bring the connection equations (Theorem 7.2 of Chapter 
II ) to bear on the study of a surface M in E 3 . Let E h E 2 , E 3 be an adapted 
frame field on M. By moving each frame Ei(p), E 2 (p), E 3 (p) over a short 




FIG. 6.3 



248 GEOMETRY OF SURFACES IN E 3 [Chap. VI 

interval on the normal line at each point p, we can extend the given frame 
field to one defined on an open set in E 3 . Thus the connection equations 

are available for use. We shall apply them only to vectors v tangent to M. In 
particular, the connection forms w»y become 1-forms on M in the sense of 
Section 4 of Chapter IV. Thus we have 

1.4 Theorem If Ei,E 2 ,E 3 is an adapted frame field on M C E 3 , and v 
is tangent to M at p, then 

V^ = £ Wt -y(v)^-(p) (1 ^ i ^ 3). 

The usual interpretation of the connection forms may be read from these 
equations, and it bears repetition: o> t -j(v) is the initial rate at which Ei ro- 
tates toward E, as p moves in the v direction. Since E 3 is a unit normal vector 
field on M, the shape operator of M can be described by connection forms. 

1.5 Corollary Let S be the shape operator gotten from E 3 , where 
Ei, Ei, E 3 is an adapted frame field on M C E 3 . Then for each tangent 
vector v to M at p, 

£(v) = W U (v).Ei(p) + C028(v)# 2 (p) 

Proof. By definition, S(\) = — V V E 3 . Thus the connection equation for 
i — 3 gives the result, since the connection form co = (tan) is skew-sym- 
metric: Uij = — C0/i. | 

In addition to its connection forms, the adapted frame field E lt E 2 , E 3 
also has dual 1-forms 8 i} 2 , 03 (Definition 8.1 of Chapter II) which give 
the coordinates t (v) = v£j(p), of any tangent vector \ p with respect to 
the frame i?i(p), ^(p), ^(p)- As with the connection forms, the dual 
forms will be applied only to vectors tangent to M, so they become forms 
on M. This restriction is fatal to 6 S , for if v is tangent to M, it is orthogonal 
to E 3 , so 3 (v) = v£i(p) = 0. Thus 3 is identically zero on M. 

Because of the skew-symmetry of the connection form, we are left with 
essentially only five 1-forms: 

0i, 2 provide a dual description of the tangent vector fields E u E 2 

ooi2 gives the rate of rotation of Ei, E 2 

^13, ^23 describe the shape operator derived from E 3 

1.6 Example The sphere. Consider the adapted frame field E lf E 2 , E 3 
defined on the (doubly punctured) sphere 2 in Example 1.3. By extending 
this frame field to an open set of E we get the spherical frame field given in 
Example 6.2 of Chapter II, provided the indices of the latter are shifted by 



Sec. 1] THE FUNDAMENTAL EQUATIONS 249 

1 — ► 3, 2 — ► 1, 3 — » 2. Thus, in terms of the spherical coordinate functions, 
Example 8.4 of Chapter II gives 

6i = r cos <p d& 2 = r d<p 

cow = sin <p d& coi3 = — cos $? d# 023 = —dtp 

Because all forms (including functions) are now restricted to the surface 
2, the spherical coordinate function p has become a constant: the radius r 
of the sphere. 

In general the forms associated with an adapted frame field obey the 
following remarkable set of equations. 

1.7 Theorem If E ly E 2 , E z is an adapted frame field onlc E 3 , then its 
dual forms and connection forms on M satisfy: 

\dd\ = OJi2 A 2 

(1)1 First structural equations 

Id02 = U21 a 6\ 



(2) a>3i a 0i + 0332 a 0-2 — Symmetry equation 

(3 ) do3i2 = o>i3 a o> 3 2 Gauss equation 

[do>i3 = OJi2 A C023 

(4)-| Codazzi equations 

|do>23 = «21 A OJ]3 



Proof. We merely apply the structural equations in Theorem 8.3 of 
Chapter II. The first structural equation 

ddi — zl Ma a dj 

i 

yields (1) and (2) above. In fact, for i = 1, 2, we get (1), since 3 = 
on the surface M. But 3 = implies d6 3 = 0, so for i = 3 we get (2). 

Then the second structural equation yields the Gauss (3) and Codazzi 
(4) equations. I 

Because connection forms are skew -symmetric, and a wedge product of 
1 -forms satisfies <j> a \p = —\f/ a <f>, the fundamental equations above can 
be rewritten in a variety of equivalent ways. However, we shall stick to 
the index pattern used above, which, on the whole, seems the easiest to 
remember. 

We emphasize that the forms introduced in this section describe, not the 
surface M directly, but only the particular adapted frame field E lf E 2 , E z 
from which they are derived : A different choice of the frame field will pro- 
duce different forms. Nevertheless the six fundamental equations in Theorem 
1.7 contain a tremendous amount of information about the surface M c E , 
and we shall call on each in turn as we come to a geometric situation that 



250 



GEOMETRY OF SURFACES IN E 3 



[Chap. VI 



it governs. For example, since o>i 3 and 0023 describe the shape operator of M, 
the Codazzi equations (4) express the rate at which the shape of M is 
changing from one point to another. 

The first of the following exercises shows how the Cartan approach 
automatically singles out the three types of curves considered in Chapter 
V, Section 5. 



EXERCISES 

1. Let a be a unit-speed curve in M C E 3 . If Ei, E 2 , E 3 is an adapted frame 
field such that E x restricted to a is its unit tangent T, show that 

(a) a is a geodesic of M if and only if a>i 2 (T) = 0. 

(b) If E 3 = EtX E 2 , then 

g = un(T), k = Wl3 (T), t = u> 23 {T) 

where g, k, and t are the functions defined in Ex. 7 of V.5. {Hint: If T = 
Ei along a, then E/ = V El Ei along a. ) 

2. {Sphere). For the frame field in Example 1.6: 

(a) Verify the fundamental equations (Theorem 1.7). 

(b ) Deduce from the formulas for 0i and 2 that 

E x [#\ = l/r cos <P E } [<p] = 
E 2 [d] = Et[<p] = l/r 

(c) Use Corollary 1.5 to find the shape operator S of the sphere. 

3. {Torus). Let E u E 2 , E 3 be the adapted frame field on the torus T (radii 
R > r) such that E 2 is tangent to meridians and E x is tangent to paral- 
lels, as in Fig. 6.4. Use the toroidal frame field in E 3 to get 

0i = (R + r cos <p) d# 

2 = r d<p 

CJ12 = sin <p d# 

o)i3 = — cos <p d& 

C0 2 3 = —dip 

Check the fundamental 
equations for these forms. 
4. {Continuation). By the 
fig. 6.4 methods of this section, 




Sec. 2] FORM COMPUTATIONS 251 

compute S(Ei) and S(E 2 ) for the frame field above. Deduce that 
meridians and parallels are principal curves, and find the principal cur- 
vature functions. (Compare Example 6.1 of Chapter V, where the unit 
normal is "inward.") 

5. Use the cylindrical frame field in E 3 (Example 6.2 of Chapter II) to 
compute the shape operator of the cylinder M: x + y = r 2 . 

6. Give a new proof that shape operators are symmetric by using the 
symmetry equation (Theorem 1.7). 



2 Form Computations 

From now on, our study of the geometry of surfaces will be carried on 
mostly in terms of differential forms, so the reader may wish to look back 
over their general properties in Sections 4 and 5 of Chapter IV. Increas- 
ingly we shall tend to compare M with the Euclidean plane E 2 . Thus if 
Ei, E 2 , E z is an adapted frame field onMc E 3 , we say that E lf E 2 con- 
stitutes a tangent frame field on M. Any tangent vector field V on M may 
be expressed in terms of Ei and E 2 by the orthonormal expansion 

V = V>EiEi+ V'E 2 E 2 

To show that two forms are equal, we do not have to check their values on 
all tangent vectors, but only on the "basis" vector fields E x , E 2 . (See the 
remarks preceding Example 4.7 of Chapter IV). Explicitly: l-forms <t> and 
yp are equal if and only if 

<j>(Ei) = +(Ei) and <t>(E 2 ) = +(E 2 ); 

2-forms n and v are equal if and only if 

niE^E^ = v(Ei,E 2 ). 

The dual forms d u 6 2 are, as we have emphasized, merely another descrip- 
tion of the tangent frame field E x , E 2 ; they are completely characterized by 
the equations 

e t (Ei) = 8 {j (1 £i,j ^2). 

These forms provided a "basis" for the forms on M (or, strictly speaking, 
on the region of definition of E u E 2 ). 

2.1 Lemma (The Basis Formulas) Let di, 6 2 be the dual l-forms of 
Ei, E 2 on M. If is a 1-form and n a 2-form, then 

(1) $ = <f>(Ei) e x +4>(E 2 ) 2 

(2) M = p(E u E 2 ) d x a 2 



252 GEOMETRY OF SURFACES IN E 3 [Chap. VI 

Proof. Apply the equality criteria above, observing for (2) that by 
definition of the wedge product, 

(0i a 6 2 )(Ex,E 2 ) = d 1 (E 1 )6 2 (E 2 ) - 6 1 (E 2 )d 2 (E 1 ) 

= 11 - 00 = 1. I 

Assuming throughout that the forms X , 2 , o> n , wu, w 23 derive as in Section 
1 from an adapted frame field Ex, E 2 , E 3 on a region in M, let us see what 
some of the concepts introduced in Chapter V look like when expressed in 
terms of forms. We begin with the analogue of Lemma 3.4 of Chapter V. 

2.2 Lemma 

(1) 0>i3 A C0 2 3 = K6i A 2 . 

(2) WW A 2 -f X A a>23 = 2#0! A 6 2 . 

Proof. To apply the definitions K = det S, 2H = trace S, we shall find 
the matrix of S with respect to E x and E 2 . As in Corollary 1.5 the connection 
equations give 

S(Ex) = -V El E s = -o} 31 (Ex)Ex - u Z2 {Ex)E 2 
S(E 2 ) = -V E2 E 3 = -u> n (E 2 )Ex - u Z2 (E 2 )E 2 . 
Thus the matrix of S is 

(wx%(Ex) uisiExY 

\wxi{E 2 ) oo 2 z(E 2 ) t 

Now, by the second formula in Lemma 2.1, what we must show is that 

(cow a comX^i, E z ) = K and (wu a 2 -f di a wa)(^i, ^2), = 2H. 
But 

(ww A u n ){Ex,E 2 ) = oixz{Ex)oi 2i {E 2 ) - Uiu{E 2 )o^{Ex) 

= determinant of matrix of S = det S = K 
and a similar computation gives the trace formula. | 

Comparing the first formula above with the Gauss equation (3) in 
Theorem 1.7, we get 

2.3 Corollary dw l2 = —Kdx a 2 . 

We shall call this the second structural equation,] and derive from it a 
new interpretation of Gaussian curvature: u n measures the rate of rotation 
of the tangent frame field E u E 2 —a,nd since K determines the exterior 
derivative dw 12 , it becomes a kind of "second derivative" of E u E 2 . 

t This equation will be shown to be the analogue for M of the second struc- 
tural equation (Theorem 8.3 of Chapter II) for E 3 . 



Sec. 2] FORM COMPUTATIONS 253 

For example, on a sphere 2 of radius r, the formulas in Example 1.6 give 

di a 2 = r cos <p d& d<p = — r 2 cos <p d<p d&. 

But 

do)^ = d(sin<pd&) = d(sin.(p) a d& = cos <p d*pdd. 

Thus the second structural equation gives the expected result, K = 1/r 2 . 
This new description of curvature may be rewritten in still another way. 

2.4 Corollary K = # 2 [a>i 2 (#i)] - #i[» u (tf 2 )] - ^{Exf - ^(E 2 )\ 
Proof. By Lemma 2.1, we have 

Wl2 = fl6i + / 2 2 , 

where 

fi = un(Ei) fori = 1, 2. 
Then 

dun = rf/l A 0! + rf/ 2 A 2 + /i d0 x + / 2 d0 2 

= d/ X A 0! + rf/ 2 A 2 + /lCO i2 A 2 + /2C02! A X 

where we have used the first structural equations (Theorem 1.7). Now 
apply this formula to E u E 2 . Since 0;(2£y) = 5 t y, we get 

douiE^Ei) = -dME 2 ) +df i (E 1 ) +fi<» a (Ei) - fa^iE*) 

Hence, using the previous corollary, 

— K= — E 2 [fi] + Ei[fi] -\- f\o)n(Ei) +/ 2 c»)i 2 (E 2 ) 

which, by definition of /i and / 2 , is the required result. | 

For instance, from Example 1.6 we readily compute that 

uniEy) = - tan <p and o)\ 2 {E 2 ) = 0. 
r 

Thus f or the sphere, the formula above yields 

2 



K = EA- tan <p — [ - tan <p J = — 

since by Exercise 2 of Section 1 we have 



— tan tp _ 1 



£"2 [tan <p] = sec 2 <p E 2 [<p] = 

T 

We emphasized in Section 1 that, in general, adapted frame fields on 
M (Z E give only indirect information about M. If such a frame field is to 



254 GEOMETRY OF SURFACES IN E 3 [Chap. VI 

give direct geometric information, it must be derived in some natural way 
from the geometry of M itself, as was the case with the Frenet frame field 
of a curve. There is a way to do this : 

2.5 Definition A principal frame field on M C E is an adapted frame 
field Ei, E 2 , E 3 such that at each point E x and E 2 are principal vectors of M. 

So long as its domain of definition contains no umbilics, a principal 
frame field is uniquely determined — except for changes of sign — by the 
two principal directions at each point. 

Occasionally it may be possible to get a principal frame field on an entire 
surface. For example, on a surface of revolution, we can take E\ tangent to 
meridians, E 2 tangent to parallels. In general, however, about the best we 
can do is as follows. 

2.6 Lemma If p is a nonumbilic point of I c E 3 , then there exists a 
principle frame field on some neighborhood of p in M. 

Proof. Let Fi, F 2 , Fz be an arbitrary adapted frame field on a neighbor- 
hood 91 of p. Since p is not umbilic, we can assume (by rotating Fi, F 2 
if necessary) that Fi(p) and Fi (p) are not principal vectors at p. By 
hypothesis fci(p) ?± fc 2 (p); hence by continuity ki and k 2 remain distinct 
near p. On a small enough neighborhood 31 of p, all these conditions are 
thus in force. 

Let Sij be the matrix of $ with respect to F u Fi. It is now just a standard 
problem in linear algebra to compute — simultaneously at all points of 91 — - 
characteristic vectors of S, that is, principal vectors of M. In fact, at each 
point the tangent vector fields 

7i = S 12 F t + (h - S U )F 2 

V 2 = (k 2 — $22)^1 + S\ 2 F 2 

give characteristic vectors of S. (This can be checked by a direct computa- 
tion if one does not care to appeal to linear algebra.) Furthermore, the 
function $12 = S(Fi)'F 2 is never zero on our selected neighborhood 91, 
so || Vi || and || V 2 || are never zero. Thus the vector fields 

El = || J7 El = 



.. Vi 11 11 y 2 11 

consist only of principal vectors, so E u E 2 , E 3 = F 3 is a principal frame 
field on 91. | 

If Ei, Ei, Ez is a principal frame field on M, then the vector fields Ei and 
E 2 consist of characteristic vectors of the shape operator derived from E z . 
Thus we can label the principal curvature functions so that S (Ei ) = k t Ei 
and S(E 2 ) = k 2 E 2 . Comparison with Corollary 1.5 then yields 



Sec. 2] FORM COMPUTATIONS 255 

CtflsCEl) = fcl 03 U (E2) = 

um{Ex) = 0023(^2) = fa- 

Thus the basis formula (1) in Lemma 2.1 gives 

C0i3 = fadl 0>23 = fa&2 ( ) 

This leads to an interesting version of the Codazzi equations. 

2.7 Theorem If E u E 2 , E s is a principal frame field on M C E 3 , then 

EM = (fa ~ fa) u l2 (E 2 ) 

E 2 [fa] = (fa - fa)o, 12 (E l ) 
Proof. The Codazzi equations (Theorem 1.7) read 

UOliz = Wl2 A OJ23 0,0323 = W21 ^ W13. 

The proof is now an exercise in the calculus of forms as discussed in Chapter 
IV, Section 4. Substituting from (*), above, in the first of these equations, 
we get 

d(fadi) = CO12 a fad 2 ; 

hence 

dfa a di -\- fa dd]_ = fc 2 aji2 a 2 
If we substitute the structural equation dB\ = a>i 2 a 6 2 , this becomes 

dfa A 6 1 = (&2 — fa) CO12 A 2 

Now apply these 2-forms to the pair of vector fields E if E 2 to obtain 

- dfa(E 2 ) = (fa - fa) w M C&i) - 0, 
hence 

E 2 [fa] = dfa(E 2 ) = (fa - fa) ««(#!). 

The other required equation derives in the same way from the Codazzi 
equation dutx = 0021 a W13. I 

Note that for a principal frame field, uu(v) tells how the principal 
directions are changing in the v direction. 

EXERCISES 

1. Verify the Codazzi equations (Theorem 2.7) for the principal frame 
field on the torus given in Exercise 3 of Section 1. (Hint: 



12 - 



gV[f) - fV[g] 



256 GEOMETRY OF SURFACES IN E 3 [Chap. VI 

the derivatives of # and <p with respect to E x and E 2 may be found from 
the formulas for X and 2 in this exercise.) 

2. If E h E 2 , E 3 is an adapted frame field on M with E x • E 2 X E 3 = 1, let 

Mp) = P • #«(p) for i = 1, 2, 3. 

In particular, hz is the support function h of M defined on page 218. 
Show that 

dhi = 0i + (dwhi + W13/13 

<//&2 = 02 + 0)21^1 + G>23^3 

(flm£: Strictly speaking, h { = X > E iy where X is the remarkable vector 
field such that V V X = V used also on page 218.) 

There is a rough rule for computing exterior derivatives of 1 -forms in 
terms of an adapted frame field: Express the form in terms of 0i and 2 
(or perhaps &»,-,•); then apply d and use the fundamental equations. The 
proof of Theorem 2.7 is one example; another is as follows. 

3. (Continuation). 

(a) If ^ is the 1-form such that ^(v) = p • v X ^(p), show that 

# = 2(1 + hH) X a 2 , 

where h is the support function. 

(b) If f is the 1-form such that f (v) = ^(-S(v)), show that 

rff = 2(H + hK) 0i a 2 . 
(Hint: \f/ = hi 2 — /i 2 0i, and f has an analogous expression.) 



3 Some Global Theorems 

We have claimed all along that the shape operator S is the analogue for 
a surface M of the curvature and torsion of a curve in E 3 . Simple hypotheses 
on k and t singled out some important special types of curves. Let us now 
see what can be done with S in the case of surfaces. (Recall that we are 
dealing exclusively with connected surfaces. ) 

3.1 Theorem If its shape operator is identically zero, then M is (part 
of ) a plane in E 3 . 

Proof. The scheme used is analogous to that of Corollary 3.5 of Chapter 
II. By definition of the shape operator, S = means that any unit normal 
vector field E z on M is Euclidean parallel; thus it may be identified with a 
point of E 3 . Fix a point p of M. We shall show that M lies in the plane 



Sec. 3] 



SOME GLOBAL THEOREMS 



257 




FIG. 6.5 



through p orthogonal to E 3 . If q is an arbitrary point of M , then since M 
is connected, there is a curve a in M from a (0) = p to a (1 ) = q. Consider 
the function 



Now 



df 

dt 



f(t) = («(*) -p) .E». 



a*E 3 = and /(0) = 0; 



hence / is identically zero. In particular, 

/(I) = (q-p) -E 3 = 0, 
so every point q of M is in the required plane (Fig. 6.5). 



I 



We saw in Chapter V, Section 3 that requiring a single point p of M C E 3 
to be planar (&i = k 2 = 0, or equivalently *S = 0) produces no significant 
effect on the shape of M near p. But the result above shows that if every 
point is planar, then M is, in fact, part of a plane. 

Perhaps the next simplest hypothesis on a surface M in E 3 is that at 
each point p, the shape operator is merely scalar multiplication by some 
number — which a priori may depend on p. This means that M is all- 
umbilic, that is, consists entirely of umbilic points. 

3.2 Lemma If M is an all-umbilic surface in E 3 , then M has constant 
Gaussian curvature K ^ 0. 

Proof. Let E x , E 2 , E 3 be an adapted frame field on some region in M. 
Since M is all-umbilic, the principal curvature functions on are equal, 
Jd = kt = k, and furthermore E u E 2 , E 3 is actually a principal frame field 
(since every direction on M is principal). Thus we can apply Theorem 
2.7 to conclude that E x [k] = E 2 [k] = 0. Alternatively we may write 

dkiEi) = dk(E 2 ) = 0, 



258 GEOMETRY OF SURFACES IN E 3 [Chap. VI 

so by Lemma 2.1, dk = on 0. But K = kik 2 = k 2 , so dK = Ik dk = 
on 0. Since every point of M is in such a region 0, we conclude that dK = 
on all of M. It follows by an earlier exercise that K is constant. | 

3.3 Theorem If M d E 3 is all-umbilic and K > 0, then M is part of 
a sphere in E 3 of radius \/\/K. 

Proof. (This time the scheme of proof is analogous to Lemma 3.6 of 
Chapter II.) Pick at random a point p in M and a unit normal vector 
^(p) to\M at p. We shall prove that the point 

c = p + ^) Ea(p) 

is equidistant from every point of M. (Here A;(p) = fci(p) = fc 2 (p) is the 
principal curvature corresponding to 2£ 3 (p).) 

Now let q be any point of M, and let a be a curve segment in M from 
a(0) = ptoa(l) = q. Extend -E^p) to a unit normal vector field E 3 on 
a, as shown in Fig. 6.6, and consider the curve 

7 = a + r Ez in E . 
k 

Here we understand that the principal curvature function k derives 
from E 3 , thus k is continuous. But K = k 2 and by the preceding lemma, 
K is constant, so k is constant. Thus 

7 = a + — Ez . 



But 



E 3 ' = -S(a) = - ka, 
E,(p) E t 



Sec. 3] SOME GLOBAL THEOREMS 259 

since by the all-umbilic hypothesis, S is scalar multiplication by k. Thus 

7 - a + t (~ka) - 0, 
so the curve 7 must be constant. In particular 

c = 7(0) = 7(D = q+i# 3 (q) 

so <l(c, q) = 1/ I k I for every point q of M. Since K = kik 2 = k 2 , we 
have shown that M is contained in the sphere of center c and radius 

i/Vk. I 

Using all three of the preceding results, we conclude that a surface M 
in E is all-umbilic if and only if M is part of a plane or a sphere. 

3.4 Corollary A compact all-umbilic surface M in E is an entire sphere. 

Proof. By the preceding remark, we deduce from Theorem 7.6, Chapter IV, 
that M must be an entire plane or sphere. The former is impossible, since 
M is — by hypothesis — compact, but planes are not. | 

Gaussian curvature was used in the preceding results mostly because 
it is well-defined and differentiable on all of M, and is thus easier to work 
with than principal curvatures. 

We now turn to a more serious examination of the Gaussian curvature 
K of a surface M (Z E 3 . 

3.5 Theorem On every compact surface M in E 3 , there is a point at 
which the Gaussian curvature K is strictly positive. 

Proof. Consider the real-valued function/ on M such that/(p) = || p || 2 . 
Thus in terms of the natural coordinates of E 3 , / = ^2 %?• Now / is differ- 
entiable, hence continuous, and M is compact. Thus by Lemma 7.3 of 
Chapter IV, / takes on its maximum at some point m of M . Since / meas- 
ures the square of the distance to the origin, m is simply a point of M at 
maximum distance r = || m || > from the origin. Intuitively it is clear 
that M is tangent at p to the sphere 2 of radius r — and that M lies inside 
S, hence is more curved than 2 (Fig. 6.7). Thus we would expect that 
K(m) > 1/r 2 > 0. Let us prove this inequality. 

Given any unit tangent vector u to M at the maximum point m, pick 
a unit-speed curve a in M such that a(0) = m, a (0) = u. It follows 
from the derivation of m that the composite function f(a) also has a 
maximum at t = 0. Thus 

|(/a)(0)=0, |- 2 (/«)(0) ^0. (1) 



260 GEOMETRY OF SURFACES IN E 3 [Chap. VI 




FIG. 6.7 

j 
But /(a) = a*a, so -r(fa) = 2a • a. Evaluating at t = 0, we find 

= ^M. (0) = 2a(0).«'(0) = 2m.u. 
at 

Since u was any unit tangent vector to M at m, this means that m (con- 
sidered as a vector) is normal to M at m. 
Differentiating again, we get 

d (fa) n ' ' s n tf 
eft 2 

By (1), aU = this yields 

> u • u + m • a" (0) 

(2) 
= 1 + m .a"(0). 

The discussion above shows that m/r may be considered as a unit normal 
vector to M at m as shown in Fig. 6.8. Thus (m/r)»a" is precisely the 
normal curvature k(u) of M in the u direction, and it follows from (2) 
that k(u) ^ — 1/r. In particular, both principal curvatures satisfy this 
inequality, so 

K(m)^- 2 >0. | 

Thus there are no compact surfaces in E with K ^ 0. 

Maintaining the hypothesis of compactness, we consider the effect of 
requiring that Gaussian curvature be constant. Theorem 3.5 shows that 
the only possibility is K > 0. Spheres are obvious examples of compact 
surfaces in E 3 with constant positive Gaussian curvature. It is one of the 
most remarkable facts of surface theory that they are the only such sur- 
faces. To prove this we need a rather deep preliminary result. 



Sec. 3] 



SOME GLOBAL THEOREMS 

m/r 



261 




FIG. 6.8 



3.6 Lemma (Hilbert) Let m be a point of McE such that 

(1) ki has a local maximum at m. 

(2) fc 2 has a local minimum at m. 

(3) fcx(m) > k(m). 
ThenK(m) ^ 0. 

For example, it is easy to see that these hypotheses hold at any point 
on the inner equator of a torus or on the minimal circle (x = 0) of the 
catenoid. And K is, in fact, negative in both these examples. 

To convert hypotheses (1) and (2) into usable form in the proof that 
follows, we recall some facts about maxima and minima. If / is a (differ- 
entiable) function on a surface M and V is a tangent vector field, then the 
first derivative V\f] is again a function on M. Thus we can apply V again 
to obtain the second derivative F[F[/]] = FFf/]. A straightforward com- 
putation shows that if / has a local maximum at a point m, then the ana- 
logues of the usual conditions in elementary calculus hold, namely 

V[f] = 0, VV[f] SO at m. 

For a local minimum, of course, the inequality is reversed. 

Proof. Since fci(m) > /^(m), m is not umbilic, hence by Lemma 2.6 
there exists a principal frame field E it E 2 , E 3 on a neighborhood of m in 
M. By the remark above, the hypotheses of minimality and maximality 
at m imply in particular 

EM = Etlh] = atm (1) 

and 

EiEM ^ and E^lh] ^0 at m. (2) 



262 GEOMETRY OF SURFACES IN E 3 [Chap. VI 

Now we use the Codazzi equations (Theorem 2.7). From (1) it follows 
that 

oin(Ei) = con(E 2 ) =0 at m 

since fci — k 2 5* at m, thus by Corollary 2.4 

K = EJLwniEi)] - EtlaniEi)] at m. (3) 

Applying Ei to the first Codazzi equation in Theorem 2.7 yields 

EMh] = CEitfci] - E^h]) <**(E 2 ) + (fci - A*) #i[« B (#,)]. 

But at the special point m, we have o>n(E 2 ) = and fci — k 2 > 0; hence 
from (2 ) we deduce 

Ei[an(E 2 )] ^ atm. (4) 

A similar argument starting from the second Codazzi equation gives 

EtiuniEi)] S atm. (5) 

Using (4) and (5) in the expression (3) for the Gaussian curvature at m, 
we conclude that K(m) ^ 0. I 

3.7 Theorem (Liebmann) If M is a compact surface in E with constant 
Gaussian curvature K, then M is a sphere of radius l/\/K. (Theorem 3.5 
implies K is positive.) 

Proof. We do not know that M is orientable, so principal curvature 
functions are not available on all of M. Nevertheless, the function 

H 2 - K = (h- fc 2 ) 2 /4 

is well-defined and continuous on all of M, since squaring eliminates 
ambiguity of sign. Because M is compact, the function H 2 — K ^ has 
a maximum point m. Now if H 2 — K is zero at m, it is identically zero; 
thus M is all-umbilic and Corollary 3.4 gives the required result. 

So what we must show is that H 2 — K cannot be positive at m. Assume 
it is; then m is not umbilic and by suitably orienting a neighborhood 91 
of m we can arrange that h > k 2 > on 91 (since K > 0). Then ki — k? 
has a maximum at m, since (fci — k 2 ) does. Since K = kik 2 is constant, 
it follows that ki has a local maximum at m and k 2 has a local minimum. 
But now we can apply Hilbert's lemma to obtain the contradiction 

K(m) ^0. I 

Liebmann's theorem is false if the compactness hypothesis is omitted, 
for we saw in Chapter V, Section 6, that there are many nonspherical 
surfaces in E 3 with constant (positive) curvature. Both Theorem 3.5 and 
Liebmann's theorem depend on the fundamental topological fact (Lemma 



Sec. 4] ISOMETRIES AND LOCAL ISOMETRIES 263 

7.3 of Chapter IV) that a continuous real- valued function on a compact 
surface has a maximum. More advanced topological methods are required 
for a full investigation of the influence of Gaussian curvature on the shape 
of surfaces in E 3 . For example, one might ask what the situation is for 
constant curvature surfaces when compactness is weakened to closure 
in E 3 . f Here are the answers : 

A closed surface M d E 3 with K constant positive is compact — hence 
by Liebmann's theorem it is a sphere. 

A closed surface M C E 3 with K = is a generalized cylinder (Massey). 

There are no closed surfaces in E 3 with K constant negative (Hilbert). 

We shall prove the first result in Chapter VII. Proofs of the last two may 
be found in Hicks [5] and Willmore [3], respectively. 



EXERCISES 

1. If M is a flat minimal surface, prove that M is part of a plane. 

2. Flat surfaces in E 3 can be bent only along straight lines: If fa = 0, but fa 
is never zero, show that the principal curves of fa are straight-line seg- 
ments in E 3 . (Hint: Use Ex. 1 of VI. 1.) 

This is the starting point for the proof of Massey's theorem. 

3. Let I c E J be a compact orientable surface with K > 0. If M has 
constant mean curvature, show that M is a sphere. 

4. Prove that in a region free of umbilics there are exactly two principal 
curves (ignoring different parametrizations) through each point, these 
crossing orthogonally. (Hint: Use Ex. 7 of IV.8.) 

5. If the principal curvatures of a surface IcE 3 are constant, show that 
M is part of either a plane, a sphere, or a circular cylinder. (Hint: In 
the nontrivial case fa ?* fa, assume there is an adapted frame field on 
all of M, and show that, say, fa = 0.) 



4 Isometries and Local Isometries 

We referred earlier to properties of a surface M in E that could be dis- 
covered by inhabitants of M unaware of the space outside their surface. 
We asserted that the inhabitants of M could determine the distance in M 

t This condition is defined in Exercise 10 of Chapter IV, Section 7. Roughly 
speaking, it means that M has no edges or rims. For surfaces in E 3 it is equivalent 
to the intrinsic property of completeness (Definition 4.4, of Chapter VII). 



264 



GEOMETRY OF SURFACES IN E 3 



[Chap. VI 



between any two points, just as the distance along the surface of the earth 
is found by its inhabitants. The mathematical formulation is as follows. 

4.1 Definition If p and q are points of M c E 3 , consider the collection 
of all curve segments a in M from p to q. The intrinsic distance p(p, q) 
from p to q in M is the greatest lower bound of the lengths L(a) of these 
curve segments. 

There need not be a curve a whose length is exactly />(p, q) (see Exercise 
3). The intrinsic distance p(p, q) will generally be greater than the straight- 
line Euclidean distance d(p, q), since the curves a are obliged to stay in 
M (Fig. 6.9). 

On the surface of the earth (sphere of radius ca. 4000 miles) it is, of 
course, intrinsic distance that is of practical interest. One says, for example, 
that it is 12,500 miles from the north pole to the south pole — the Euclidean 
distance through the center of the earth is only 8000. 

We saw in Chapter III how Euclidean geometry is based on the notion 
of isometry, a distance-preserving mapping. For surfaces in M we shall 
prove the distance-preserving property and use its infinitesimal form 
(Corollary 2.2 of Chapter III) as definition. 

4.2 Definition An isometry F: M — ► M of surfaces in E 3 is a one-to-one 
mapping of M onto M that preserves dot products of tangent vectors. 
Explicitly, if F* is the derivative map of F, then 

F*(v)»F*(w) = vw 
for any pair of tangent vectors v, w to M. 




FIG. 6.9 



Sec. 4] ISOMETRIES AND LOCAL ISOMETRIES 265 

If F* preserves dot products, then it also preserves lengths of tangent 
vectors. It follows that an isometry is a regular mapping (Chapter IV, 
Section 5), for if F* (v) = 0, then 

||v|| = ||F«(y)|| =0, 

hence v = 0. Thus by the remarks following Theorem 5.4 of Chapter IV, 
an isometry F: M — ► M is in particular a diffeomorphism, that is, has an 
inverse mapping F -1 : M — > M. 

4.3 Theorem Isometries preserve intrinsic distance: if F: M — > M is 
an isometry of surfaces in E 3 , then 

p(p,q) =iJ(F(p),F(q)) 

for any two points p, q in M. 

(Here p and p are the intrinsic distance functions of M and M respec- 
tively.) 

Proof. First note that isometries preserve the speed and length of curves. 
The proof is just like the Euclidean case: If a is a curve segment in M, 
then a = F(a) is a curve segment in M with velocity a' = F*(a'). Since 
F* preserves dot products, it preserves norms, so || a' || = || F*(a') |J = 
|| F(a)' || = || a' || . Hence 

L(a) = f || a' (t) \\dt= f || a'(t) || dt = L(a). 

•'o J a 

Now if a runs from p to q in M, its image a = F(a) runs from F(p) 
to F(q) in M. Reciprocally, if is a curve segment in M from ^(p) to F(q) 
in M, then F _1 (/8) runs from p to q in M. We have in fact established a 
one-to-one correspondence between the collection of curve segments used 
to define p(p, q) and those used for p(F(p), F(q)). But as shown above, 
corresponding curves have the same length, so it follows immediately 
that p(p, q) = p(F(p), F(q)). I 

Thus we may think of an isometry as bending a surface into a different 
shape without changing the intrinsic distance between any of its points. 
Consequently the inhabitants of the surface are not aware of any change at 
all, for their geometric measurements all remain exactly the same. 

If there exists an isometry from M to M, then these two surfaces are 
said to be isometric. For example, if a piece of paper is bent into various 
shapes without creasing or stretching, the resulting surfaces are all iso- 
metric (Fig. 6.10). 

To study isometries it is convenient to separate the geometric condition 
of preservation of dot products from the one-to-one and onto requirements. 

4.4 Definition A local isometry F: M — + N of surfaces is a mapping that 
preserves dot products of tangent vectors (that is, F* does). 



266 



GEOMETRY OF SURFACES IN E 3 



[Chap. VI 






FIG. 6.10 



Thus an isometry is a local isometry that is one-to-one and onto. 

If F is a local isometry, the earlier argument still shows that F is a regular 
mapping. Then for each point p of M the inverse function theorem (5.4 
of Chapter IV) asserts that there is a neighborhood It of p in M that F 
carries diffeomorphically onto a neighborhood V of F(p) in N. Now "U 
and V are themselves surfaces in E 3 , and thus the mapping F | It: 11 — > V 
is an isometry. In this sense a local isometry is, indeed, locally an isometry. 

There is a simple patch criterion for local isometries using the functions 
E, F, and G defined in Section 4 of Chapter V. 

4.5 Lemma Let F: M — * N be a mapping. For each patch x: D — > M, 
consider the composite mapping 

x = F(x):D^N. 

Then F is a local isometry if and only if for each patch x we have 



E = E 



F = F 



G = G 



(Here x need not be a patch, but E, F, and G are defined for it as usual. ) 

Proof. Suppose the criterion holds — and only for enough patches to cover 
all of M. Then by one of the equivalences in Exercise 1, to show that F* 
preserves dot products we need only prove that 



x u || = ||F*(x„) 



x u 'x v = F*(x u )-F*(x v ), 



x-ll = \\F*M 



But as we saw in Chapter IV, it follows immediately from the definition 
of F* that ^*(x M ) = x„ and F*(x„) = x v . Thus the equations above fol- 
low from the hypotheses E = E, F = F, G = G. Hence F is a local iso- 
metry. 

Reversing the argument, we deduce the converse assertion. | 

This result can sometimes be used to construct local isometries. In the 

simplest case, suppose that M is the image of a single patch x: D — > M . 



Sec. 4] ISOMETRIES AND LOCAL ISOMETRIES 267 

Then if y: D -> N is a patch in another surface, define a mapping F:M-*N 
by 

F(x(u,v)) = y(M,y) for (w,v) in D. 

If 

E = E,F = F,G = G, 

then by the above criterion, F is a local isometry. 

4.6 Example 

(1) Local isometry of a plane onto a cylinder. The plane E 2 may be con- 
sidered as a surface, with natural frame field Ui, U*. If x: E — > M is a 
parametrization of some surface, then Exercise 1 shows that x is a local 
isometry if 

x*(U t )'**(Ui) = Ui-Uj, 

Since x* (t/i) = x„, x* (C7 2 ) = x„, and U^Uj = «.-/, this is the same as re- 
quiring E = 1, F = 0, G = 1. 

To take a concrete case, the parametrization 



x(u, v) = (r cos f - J , r sin ( -j , vj 



of the cylinder M: x + y 2 = r 2 has E = 1, F = 0, G = 1. Thus x is a 
local isometry which wraps the plane E 2 around the cylinder, with hori- 
zontal lines going around cross-sectional circles, and vertical lines to rulings 
of the cylinder. 

(2) Local isometry of a helicoid onto a catenoid. Let H be the helicoid which 
is the image of the patch 

x(u,v) = (u cos v, u sin v, v). 

Furnish the catenoid C with the canonical parametrization y: E 2 — ► C 
discussed in Example 6.4 of Chapter V. Thus 

y(u,v) = (g(u), h(u) cosy, h(u) sin v) 
g(u) = sinh -1 w h(u) = \/l + u 2 . 
Let F: H — » C be the mapping such that 

F(x («,»)) = y(u,v). 
To prove that F is a local isometry, it suffices to check that 

E = 1 = E, F = = F, G = 1 + u 2 = h 2 = G. 
F carries the rulings (v constant) of H onto meridians of the surface of 
revolution C, and wraps the helices (u constant) of H around the parallels 



268 



GEOMETRY OF SURFACES IN E 3 



[Chap. VI 



of C. In particular, the central axis of H (z axis) is wrapped around the 
minimal circle x = of C. 

Figure 6.11 shows how a sample strip of H is carried over to C. 

Suppose that the helicoid H (or at least a finite region of it) has been 
stamped like an automobile fender out of a flexible sheet of steel— the 
patch x does this. Then H may be wrapped into the shape of a catenoid 
with no further distortion of the metal (Exercise 5 of Section 5). 

A similar experiment may be performed by cutting a hole in a ping-pong 
ball representing a sphere in E 3 . Mild pressure will then deform the ball 
into various nonround shapes, all of which are isometric. For arbitrary 
isometric surfaces M and M in E 3 , however, it is generally not possible 
to bend M (through a whole family of isometric surfaces) so as to produce 
M. 

There are special types of mappings other than (local) isometries that 
are of interest in geometry. 

4.7 Definition A mapping of surfaces F: M — > N is conformal provided 
there exists a real-valued function X > on M such that 

\\F*M\\ =X(p)||v p || 

for all tangent vectors to M. The function X is called the scale factor of F. 

Note that if F is a conformal mapping for which X has constant value 1, 
F is a local isometry. Thus a conformal mapping is a generalized isometry 





FIG. 6.11 



Sec. 4] ISOMETRIES AND LOCAL ISOMETRIES 269 

for which lengths of tangent vectors need not be preserved — but at each 
point p of M the tangent vectors at p all have their lengths stretched by 
the same factor. 

The criteria in Lemma 4.5 and in Exercise 1 may easily be adapted 
from isometries to conformal mappings by introducing the scale factor 
(or its square). In Lemma 4.5, for example, replaced = E by E\ 2 (x) = E, 
and similarly for the other two equations. 

An essential property of conformal mappings is discussed in Exercise 8. 



EXERCISES 

1. If F: M — > N is a mapping, show that the following conditions on its 
derivative map at one point p are logically equivalent : 

(a) F* preserves inner products. 

(b) F* preserves lengths of tangent vectors, that is, || F* (v) || = || v || 
for all v at p. 

(c ) F* preserves frames : If ei, e^ is a tangent frame at p, then 

F*M,F*M 

is a tangent frame at F(p). 

(d) For some one pair of linearly independent tangent vectors v and 
w at p 

II *■*(▼) II = II v ||, ||F*(w) || - || w ||, and F*(v).F*(w) = vw 

[Hint: It suffices, for example, to prove that (a) ==> (c) => (d) ==» (b) => (a).] 

These are general facts from linear algebra; in this context they provide 
useful criteria for F to be a local isometry. 

2. Show that each of the following conditions is necessary and sufficient 
for F: M — > N to be a local isometry. 

(a) F preserves the speeds of curves: || F(a)' || = \\ a || for all curves 
a in M. 

(b) F preserves lengths of curves: L(F(a)) = L(a) for all curve 
segments am. M. 

3. Let M be the xy plane in E 3 with the origin removed. Show that the 
intrinsic distance from ( — 1,0,0) to (1,0,0) in M is 2, but that there 
is no curve segment in M which joins these points and has length 2. 
(Hint: Ex. 11 of II. 2.) 

4. Formulate precisely and prove: Local isometries can shrink but not 
increase intrinsic distance. 



270 GEOMETRY OF SURFACES IN E 3 [Chap. VI 

5. Let a,/3 : / ■—> E 3 be unit-speed curves with the same curvature function 
k > 0, and assume that the ruled parametrization 

x(u,v) = a(u) + vT(u) 

of the tangent surface of a is actually a patch. Find a local isometry 
from the tangent surface of a to : 

(a) The tangent surface of 0. 

(b) A region D in the plane. 
(Hint: Ex. 9 of III. 5.) 

6. Show that the preceding exercise applies to the tangent surface of a 
helix and find the image region D in the plane. 

7. Modify the conditions in Exercise 1 so that they provided criteria 
for F to be a conformal mapping. Then prove that a patch x: D —* M 
is a conformal mapping if and only if E = G and F = 0. 

8. Show that a conformal mapping F: M — > N preserves angles in this 
sense: If # is an angle (0 ^ # ^ x) between v and w at p, then # is 
also an angle between F* (v) and F# (w) at F(p). 

9. If F: M — » Af is an isometry, prove that the inverse mapping 
i* 1-1 : M — > M is also an isometry. If F: M — > N and G: N -+ P are 
(local) isometries, prove that the composite mapping GF: Af — » P is a 
(local) isometry. 

10. Let x be a parametrization of all of M, x a parametrization in N. 
If F: M — > iV is a mapping such that F(x(u, v)) = x(/(w), fif(y)), then 

(a) Describe the effect of F on the parameter curves of x. 

(b) Show that F is a local isometry if and only if 

(In the general case, / and gr are functions of both u and y, and this 
criterion becomes more complicated.) 

(c) Find analogous conditions for F to be a conformal mapping. 

11. Let M be a surface of revolution, and let F: H — > M be a local isometry 
of the helicoid which (as in Example 4.6) carries rulings to meridians, 
and helices to parallels. Show that M must be a catenoid. (Hint: Use 
Exercise 10.) 

12. Let M be the image of a patch x with E = 1, F = 0, and G a function 
of u only (G v = 0). If the derivative d(s/G)/du is bounded, show 
that there is a local isometry of M into a surface of revolution. 

Thus any small enough region in M is isometric to a region in a 
surface of revolution. 



Sec. 5] INTRINSIC GEOMETRY OF SURFACES IN E 3 271 

13. Let x be the geographical patch in the sphere 2 of radius r (Example 
2.2 of Chapter IV). Stretch x in the north-south direction to produce 
a conformal mapping. Explicitly, let 

y(u,v) = x(u,g(v)) with 0(0) = 0, 

and determine g so that y is a conformal. Find the scale factor of y 
and the domain D such that y(D) omits only a semicircle of 2. (Mer- 
cator's map of the earth derives from y: its inverse is Mercator's pro- 
jection.) 

14. Show that stereographic projection P: S -> E 2 (Example 5.2 of Chapter 
IV) is conformal, with scale factor 

15. Let M be a surface of revolution whose profile curve is not closed, 
hence has a one-to-one parametrization. Find a conformal mapping 
F: M — > E 2 such that meridians go to lines through the origin and 
parallels go to circles centered at the origin. 



5 Intrinsic Geometry of Surfaces in E 3 

In Chapter III we defined Euclidean geometry to consist of those concepts 
preserved by Euclidean isometries. The same definition applies to surfaces: 
The intrinsic geometry of M C E 3 consists of those concepts — called iso- 
metric invariants — that are preserved by all isometries F: M — > M. For 
example, Theorem 4.3 shows that intrinsic distance is an isometric in- 
variant. We can now state Gauss's question (mentioned in the start of 
the chapter) more precisely: Which of the properties of a surface M in E 
belong to its intrinsic geometry? The definition of isometry (Definition 4.2) 
suggests that isometric invariants must depend only on the dot product 
as applied to tangent vectors to M. But the shape operator derives from a 
normal vector field, and the examples in Section 4 show that isometric 
surfaces in E 3 can have quite different shapes. In fact, these examples 
provide a formal proof that shape operators, principal directions, principal 
curvatures, and mean curvature definitely do not belong to the intrinsic 
geometry of M <Z E . 

To build a systematic theory of intrinsic geometry, we must look back 
at Section 1 and see how much of our work there is intrinsic to M . Using 
the dot product only on tangent vectors to M , we can still define a tangent 
frame field E u E 2 on M. Thus from an adapted frame field we can salvage 
the two tangent vector fields E u E 2 — and hence also their dual 1 -forms 



272 



GEOMETRY OF SURFACES IN E 3 



[Chap. VI 



0i, 2 . It is somewhat surprising to find that these completely determine 
the connection form a>i 2 . 

5.1 Lemma The connection form o) U = — co 2 i is the only 1-form that 
satisfies the first structural equations 

dd\ = CO12 A 2 ddi = CO21 a 1# 

Proof. Apply these equations to the tangent vector fields E h E 2 . Since 
6i{Ej) = d ih the definition of wedge product (Definition 4.3 of Chapter 
IV) gives 

"12 (#1) = dd^Ex.Et) 

(aa(E t ) = -co2i(# 2 ) = dd 2 (E 1 ,E i ). 

Thus by Lemma 2.1, a>i 2 = — «2i is uniquely determined by 0i, 2 . | 

5.2 Remark In fact, this proof shows how to construct 0012 = — a> 2 i 
without the use of Euclidean covariant derivatives (as in Section 1). 
Given Ei, E2 and thus 0i, 2 ; take the equations in the above proof as the 
definition of a>i 2 on E\ and E2. Then the usual linearity condition 

"12(F) = coi 2 (yi£'i + V2E2) = Via)i2(Ei) + V20in(,E2) 

makes u>n a 1-form on M, and one can easily check (by reversing the argu- 
ment above) that coi 2 = — «2i satisfies the first structural equations. 

If F: M — > M is an isometry, then we can transfer a tangent frame field 
Ei, E 2 on M to a tangent frame field E u E 2 on M: For each point q in M 
there is a unique point p in M such that F(p) = q. Then define 

£i(q) = nWp)) 

^ 2 (q) = F*(E 2 (p)). 



(Fig. 6.12). 




FIG. 6.12 



Sec. 5] INTRINSIC GEOMETRY OF SURFACES IN E 3 273 

In practice we shall abbreviate these formulas somewhat carelessly to 
Ei = F*{E\), E 2 = F*{E 2 ). 
Because F* preserves dot products, Ei, E 2 is a frame field on M, since 
EfEi = F*(Ei).F*(Ei) = EfEj = 5 iy . 

5.3 Lemma Let F: M — » M be an isometry, and let Ei, E2 be a tangent 
frame field on M. If E u E 2 is the transferred frame field on M, then 

(1) e x = F*(e x ), e 2 = F*(e 2 ) 

(2) co 12 = F*(a 12 ). 

Proof. (1) It suffices to prove that 0* and F*(di) have the same value on 
Ei and ^2. But for 1 ^ i,j S 2 we have 

F*(9i)(Ej) = &(F*0,-) = 0~i(#;) = «« = BiiEi). 

(2) Consider the structural equation ddi = W12 a 2 on M. If we apply 
F*, then by the results in Chapter IV, Section 5, we get 

d(F*6 1 ) = F*(ddi) = F*(«u) a F*(d 2 ). 

Hence, by (1), we have 

dOx = F*{u n ) a 2 . 

The other structural equation 

G?0 2 = C021 A dl 

gives a corresponding equation, so 

ddi = F*(w 12 ) a 2 

d6 2 = F*(wa) a 0l 

But now (2) is an immediate consequence of the uniqueness property 
(Lemma 5.1), since 

F*(«a) = F*(-«w) = -F*(«w). I 

From this rather routine lemma we easily derive a proof of the celebrated 
theorema egregium of Gauss. 

5.4 Theorem Gaussian curvature is an isometric invariant; that is if F: 
M —> M is an isometry, then 

K(p) = K(F(p)) 
for every point pinM. 

Proof. For an arbitrary point p of M, pick a tangent frame field E h E 2 



274 GEOMETRY OF SURFACES IN E 3 [Chap. VI 

on some neighborhood of p and transfer via F* to E if E 2 on M. By the 
previous lemma, F*(wn) = co x2 . According to Corollary 2.3, we have 

doiu = —Kdi a 02. 

Apply F* to this equation. By the results in Chapter IV, Section 5, we get 

d(F*«! 2 ) = F*(dun) = -F*(K)F*(d 1 ) a F*(d 2 ) 

where F*(K) is simply the composite function K(F). Thus by the pre- 
ceding lemma, 

dwi 2 = —K(F)6 1 a 2 . 

Comparison with dw 12 = — Kdi a 2 yields K = K(F); hence, in particular, 
K(p) = K(F(p)). | 

Gauss's theorem is one of the great discoveries of nineteenth-century 
mathematics, and we shall see in the next chapter that its implications are 
far-reaching. The essential step in the proof is the second structural equa- 
tion 

do)\2 = — Kdi a 6%. 

Once we prove Lemma 5.1, all the ingredients of this equation, except K, 
are known to derive from M alone — thus K must also. This means that 
the inhabitants of M CI E can determine the Gaussian curvature of their 
surface even though they cannot generally find S and have no conception 
of the shape of M in E 3 . 

The machinery of differential forms puts this heuristic reasoning beyond 
doubt by supplying the formal proof of isometric invariance in Theorem 
5.4. This remarkable situation is perhaps best illustrated by the formula 
K = kik-2,: An isometry need not preserve the principal curvatures, nor 
their sum, but it must preserve their product. Thus the shapes which isometric 
surfaces may have — although possibly quite different — are by no means 
unrelated. 

A local isometry is, as we have shown, an isometry on all sufficiently 
small neighborhoods. Thus it follows from Theorem 5.4 that local isometries 
preserve Gaussian curvature. For example, in Example 4.6 the plane and 
the cylinder both have K = 0. (This is why we did not hesitate to call the 
curved cylinder "flat". Intrinsically it is as flat as a plane.) In the second 
part of Example 4.6, at corresponding points 

x(u,v) and F(x(u,v)) = y(u,v), 

the helicoid and catenoid have exactly the same Gaussian curvature: 
-1/(1 + uf (see Examples 4.3 and 6.4 of Chapter V.) 



Sec. 5] INTRINSIC GEOMETRY OF SURFACES IN E 3 275 

Gauss's theorema egregium can obviously be used to show that given 
surfaces are not isometric. For example, there can be no isometry of the 
sphere 2 (or even a very small region of it) onto part of the plane, since 
their Gaussian curvatures are different. This is the dilemma of the map 
maker: The intrinsic geometry of the earth's surface is misrepresented by 
any flat map. 

The next section is computational; Section 7 will provide more isometric 
invariants. 



EXERCISES 

1. Geodesies belong to intrinsic geometry: If a is a geodesic in M, and 
F: M — » N is a (local) isometry, then F(a) is a geodesic of N. (Hint: See 
Ex. lof VI.l.) 

2. Use Exercise 1 to derive the geodesies of the circular cylinder (Example 
5.8 of Chapter V). Generalize to an arbitrary cylinder. 

3. For a (connected) surface, the values of its Gaussian curvature fill an 
interval. If there exists a local isometry of M onto N (in particular if 
M and N are isometric), show that M and N have the same curvature 
interval. Give an example to show that the converse is false. 

4. Prove that no two of the following surfaces are isometric: sphere, torus, 
helicoid, circular cylinder, saddle surface. 

5. Bending of the helicoid into the catenoid (4.6). For each number t in the 
interval ^ t ^ ir/2, let x t : E 2 — » E be the mapping such that 

x t (u,v) = cos t(Sc, Ss, v) + sin t( — Cs, Cc, u) 

where C = cosh u, S = sinh u, c = cos v, and s = sin v. 

Now x is a patch covering the helicoid, and x x/2 is a parametrization 
of the catenoid — these are mild variants of our usual parametrizations, 
and the catenoid now has the z axis as its axis of rotation. If we imagine 
t to be time, then x t for ^ t ^ x/2 describes a bending of the helicoid 
Mo in space which carries it onto the catenoid M^/2 through a whole 
family of intermediate surfaces M t = x t (E 2 ). Prove 

(a) M t is a surface. (Show merely that x t is regular.) 

(b) M t is isometric to the helicoid M if t < ir/2. (Show that F t : M — > M t 
is an isometry, where 

F t (x (u,v)) = x t (u,v). 

Also show that for t — ir/2, F T ft is a local isometry. ) 

(c) Each M t is a minimal surface. (Compute x uu + x vv = 0.) 



276 GEOMETRY OF SURFACES IN E 3 [Chap. VI 

(d) Unit normals are parallel on orbits: Along the curve t — > x t {u, v) by 
which the point x (u, v) of M moves to M T/2 , the unit normals U t 
of successive surfaces are parallel. 

(e) Gaussian curvature is constant on orbits. (Find K t (x t (u,v)), where 
K t is the Gaussian curvature of M t ). 

A brilliant series of illustrations of this bending is given in Struik [6]. 

6. Show that every local isometry of the helicoid H to the catenoid C must 
carry the axis of H to the minimal circle of C, and the rulings of H to 
the meridians of C, as in Example 4.6. (Compare Ex 11 of VI.4.) 



6 Orthogonal Coordinates 

We have seen that the intrinsic geometry of a surface M C E 3 may be 
expressed in terms of the dual forms 6 ly 2 , and connection form o>i 2 derived 
from a tangent frame field E u E 2 . These forms satisfy 
the first structural equations: 

dd\ = Wi2 A 6 2 

dd 2 = a>2i A $i 
the second structural equation: 

dun = —Kd\ a 02 

In this section we develop a practical way to compute these forms — and 
hence a new way to find the Gaussian curvature of M. 

The starting point is an orthogonal coordinate patch x: D — » M, one for 
which F = x u 'x v = 0. Since x u and x„ are orthogonal, dividing by their 
lengths || x„ || = \/E and || x„ || = y/G will produce frames. 

6.1 Definition The associated frame field Ei, E% of an orthogonal patch 
x: D — > M consists of the orthogonal unit vector fields E\ and E 2 whose 
values at each point x(u, v) of x(D) are 

x„(w, v)/\/E{u,v) and x v (u,v)/\/G(u,v). 

In Exercise 9 of Section 4 of Chapter IV, we associated with each patch 
x the coordinate functions u and v, which assign to each point x(u, v) the 
numbers u and v, respectively. For example, for the geographical patch x 
of Example 2.2 of Chapter IV, the coordinate functions are the longitude 
and latitude functions on the sphere 2. In the extreme case when x is the 
identity map of E 2 , the coordinate functions are just the natural coordinate 
functions (u, v) — > u, (u,v) — > v on E . 



Sec. 6] ORTHOGONAL COORDINATES 277 

For an orthogonal patch x with associated frame field E\, E%, we shall 
express 6 X , 2 , and coi 2 in terms of the coordinate functions u, v. Since x is 
fixed throughout the discussion, we shall run the risk of omitting the in- 
verse mapping x -1 from the notation. With this convention, the coordinate 
functions u = w(x -1 ) and y = f(x _1 ) are written simply u and v, and 
similarly x u and x„ now become tangent vector fields on M itself. Thus the 
associated frame field of x has the concise expression 

Er = x u /\/E E 2 = x v /VG. (1) 

Now the dual forms 0i, 2 are characterized by di(Ej) = da, and in the 
exercise referred to above it is shown that 

du(x u ) = 1 dv(x u ) = 

du(x v ) = dv(x v ) = 1. 

Thus we deduce from (1) that 

0i = y/Edu d 2 = y/Gdo. (2) 

By using the structural equations we shall find analogous formulas for 
W12 and K. Recall that for a function /, df = f u du + f v dv, where the sub- 
scripts indicate partial derivatives. Hence from (2) we get 

dB x = d(VE) a du = (VE)v dv du = ~~(vE) v ^ a ^ 

VG 

d6 2 = d(\/G) a dv = (VG)u du dv = ~WG)u dv A ^ 

\/E 

where we have used the alternation rule for wedge products and substituted 
dv = 2 /"\/(r and du = di/y/E from (2 ) . Comparison with the first struc- 
tural equations dd\ = wi 2 a 2 and dd 2 = — coi 2 a di shows that 

m2 = ^h^ du + (VG), dv (3 ) 

Vg Ve 

The logic is simple: By the computations above, this form satisfies the 
first structural equations; hence by uniqueness (Lemma 5.1), it must be W12. 

6.2 Example Geographical coordinates on the sphere. For the geographi- 
cal patch x in the sphere 2 (Example 2.2 of Chapter IV), we have com- 
puted E = r 2 cos 2 v,F = 0,G = r 2 . Thus by formula (2) above, 

di = r cos v du 2 = r dv. 

Now (\/E) v = — r sin v and {\/G)u = 0; hence by (3), 

wi 2 = sin v du. 



278 GEOMETRY OF SURFACES IN E 3 [Chap. VI 

The associated frame field of this patch is the same one we got in Ex- 
ample 1.6 from the spherical frame field in E 3 . With the notational shift 
u —> &, v — > <p, the forms above are (necessarily) also the same. But now 
we have a simple way to compute them directly in terms of the surface 
with no appeal to the geometry of E 3 . 

Finally we derive a new expression for the Gaussian curvature. In this 
context, exterior differentiation of o>i 2 as given in (3 ) yields 

du n = — ((\/E) v /\^G)v dv du + ((\/G)u/\/E) u du dv. 

From (2) we get 

0i a 2 = y/EG du dv; 

hence 

1 

— dvdu = du dv = ~T^p 0i A 02- 

Thus the formula above becomes 

dd)12 = 



vm {(w"). + (w)J * A 6l 



We now compare this with the second structural equation, 

cfai2 = — K6x a 2 . 

6.3 Lemma If x: D — » M is an orthogonal patch, then the Gaussian 
curvature K is given in terms of x by 



K = 



veg \\ y/E )u\ vg a; 



By contrast with the formula for K in Corollary 4.1 of Chapter V, the 
functions £, m , and « (which describe the shape operator) no longer appear. 
Indeed since K is now expressed solely in terms of E, F, and G, we have, 
using Lemma 4.5, another proof of the isometric invariance of Gaussian 
curvature. 

In applications it is generally easier to repeat the derivation of Lemma 
6.3 in each case, rather than look it up and substitute in it. For example, 
consider the polar parametrization x(u,v) = (u cos v, u sin v ) of the Euclid- 
ean plane E 2 . Here E = 1, F = (so x is orthogonal), and G = u. Thus 
by (2), 0i = du and 2 = u dv. Since 



Sec. 6] 



ORTHOGONAL COORDINATES 



279 



ddi = and ddi = du dv = —dv a lt 

we find cow = dv. But then do>i 2 = 0, which shows again that E 2 is flat. 

6.4 Example The natural frame field of a surface of revolution. For 
a canonical parametrization 

x(u, v) = (g{u), h(u) cos v, h(u) sin v) 

of a surface of revolution, the associated frame field has E x in the direction 
of the meridians and E* in the direction of the parallels (Fig. 6.13). Since 
x is orthogonal, with E = 1, G = h 2 , we get 0i = du and 2 = h dv. Here 
/i is a function of w alone, so h v = 0, and h u is the ordinary derivative h . 
From (3) — or by direct computation — o>i 2 = h dv, so 

h" 
dwi2 = h" du dv = — 0i a 2 

We conclude that the Gaussian curvature is K = -h"/h, in agreement with 
the results of Lemma 6.3 of Chapter V. 



EXERCISES 

1. Compute the dual forms, connection form, and Gaussian curvature for 
the associated frame field of the following orthogonal patches: 

(a) \(u, v) = (u cos v, u sin v, bv), helicoid 

(b) x(u,v) = (u cos v, u sin v, u 2 /2), paraboloid of revolution. 

(c) x(u,v) = (u cos v, u sin v, au), cone 

2. Let the patch x: D — » M be a conformal mapping. (The associated 
coordinate system is said to be isothermal.) Prove: 

(a) K = —A (log E)/2E, where A is the Laplacian: A/ = f uu + /„ . 




FIG. 6.13 



280 



GEOMETRY OF SURFACES IN E 3 

En 



[Chap. VI 




Ex = 



FIG. 6.14 



(b) The mean curvature H is zero if and only if x uu + x„„ = 0. (Hint- 
Use Ex. 7 of VI.4.) 

3. For a patch with E = G = 1, show that K = — ^ uv /sm #, where & is 
the coordinate angle. (Hint: For the frame field with Ex = x u as in 
Fig. 6.14, show that X = du + cos $ dv, 2 = sin & dv.) 

A. If x is a principal patch (Ex. 9 of V.4), prove 



(a) 



0013 = 



Ve 



du 



V'G 



(b) u = HE V 



= HG U . 



5. By refining the argument in the text, show that equations (2) and (3) 
are literally true provided d u 2 , and con are replaced by their coordi- 
nate expressions 

x*(0i), x*(0 2 ), x*(coi 2 ). 



7 Integration and Orientation 

The main aim of this section is to define the integral of a 2-form over a 
compact oriented surface. This notion does not involve geometry at all; 
it belongs to the integral calculus on surfaces (Chapter IV, Section 6). 
However, we shall motivate the definition by considering some related 
geometric problems. 

Perhaps the simplest use of double integration in geometry is in finding 
the area of a surface. To discover a proper definition of area, we start with 
a patch x: D — > M and ask what the area of its image x(D) should be. Let 
AR be a small coordinate rectangle in D with sides Am and Av. Now x 
distorts AR into a small curved region x(AR) in M, marked off in an obvious 
way by four segments of parameter curves, as shown in Fig. 6.15. 



Sec. 7] 



INTEGRATION AND ORIENTATION 



281 




(AR) 



FIG. 6.15 

We have seen that the segment from x (w, ») to x(m + Au, v) is linearly 
approximated by Au x u (evaluated at (u, v)), and the one from x(u, v) to 
x(u, v + Av) by Ay x„. Thus the region x(A.R) is approximated by the 
parallelogram in T X ( U<v) (M) with these vectors as sides. From Chapter II, 
Section 1, we know that this parallelogram has area 

|| Aux u X Avx v || = || x u X x„ || AuAv = s/EG — F 2 AuAv 

We conclude that the area of x(AR) should be approximately s/EG — F 2 
times the area AuAv of AR. At each point (u, v) the familiar expression 
■\/EG — F 2 gives the rate at which x is expanding area at (u, v). Thus it is 
natural to define the area of the whole region x(D) to be 



fj VEG - F 2 du dv 



Such integrals, of course, may well be improper; we shall avoid this 
difficulty by modifying the notion of patch. 

7.1 Definition The interior R° of a rectangle R: a ^ u ^ b, c ^ v S d 
is the open set a < u < b, c < v < d. A 2-segment x: R — » M is patchlike 
provided the restricted mapping x: R° — * M is a patch in M. 

The remark preceding Lemma 7.3 of Chapter IV shows that the area of 
x(R) is finite, since it implies that -\/EG — F 2 ^ is bounded on R. 

A patch-like 2-segment x: R — » M need not be one-to-one on the bound- 
ary of R, so its image may not be very rectangular. In fact we shall now 
see that the area of an entire compact surface may often be computed by 
covering it with a single 2-segment. 

7.2 Example Areas of surfaces 

(1) The sphere S of radius r. If the formula defining the geographical 
patch is applied to the rectangle R: —t < u < w, — tt/2 < v < ir/2, we 
obtain a 2-segment that covers the whole sphere. Now 

E = r 2 cos 2 v, F = 0, and G = r\ 



282 



GEOMETRY OF SURFACES IN E 3 



[Chap. VI 



SO 



VEG — F 2 = r 2 cos v, 
and the area of the sphere is thus 



• IT -t/2 



/ r 2 cos v du dv = 47ir 2 

it J-t/2 



(2) Torus 7 1 of radii R > r > 0. From Example 2.6 of Chapter IV we 
can derive a patchlike 2-segment covering the torus. Here 



VEG - F 2 = r(R + r cos u), 



so the area is 



A{T) = I f r(R + r cos u) du dv = ItRt 

(3) The bugle surface (Example 6.6 of Chapter V). Every surface of 
revolution M has a canonical parametrization with E = 1, F = 0, G = h . 
On a rectangle jR:a^w^6,0^w^ 2ir, x is a patchlike 2-segment 
whose image is the region of M between the parallels u = a and u = b 
(Fig. 6.16). Thus the area of this region is 

A ah = I I h du dv = 2t I h du. 

J a •'0 J a 

For the bugle surface, we saw in Chapter V that h(u) = ce~ ufc ; hence 
A a b = 2irc I e u,c du 

•'o 

c\ 2/ —ale — b/c\ 




FIG. 6.16 



Sec. 7] INTEGRATION AND ORIENTATION 283 




FIG. 6.17 



To find the area of the whole bugle — a noncompact surface — we expand 
this region, letting a — > and b — » « . 

Thus (see Remark 7.6) the bugle has finite area 



lim A a b = 2irc 

b-»oo 



To define the area of a complicated region we shall not try to fit a single 
patchlike 2-segment onto it. Instead we follow the usual scheme of ele- 
mentary calculus and break the region into simple pieces, then add their 
areas. 

7.3 Definition A paving of a region (P in M consists of a finite number of 
patchlike 2-segments xi, • • • , x k whose images fill (P in such a way that 
each point of (P is in at most one set x t - (Ri° ) . 

In short, the images of the x/s cover <P exactly and overlap only on their 
boundaries (Fig. 6.17). 

Not every region is pavable; since pavings are finite, compactness 
is certainly necessary (Definition 7.2 of Chapter IV). It is safe to assume 
that a compact region is pavable if its boundary consists of a finite number 
of regular curve segments. In particular, an entire compact surface is always 
pavable. t The area of a pavable region (P is defined to be the sum of the areas 
of xiCRi), • • • , Hk(Rk) for a paving of (P. (The consistency problem here 
will be discussed following the analogous definition, 7.5.) 

The preceding exposition shows that computation of area does not de- 
mand differential forms, but integration of 2-forms (Definition 6.3 of 
Chapter IV) will give area — and much more besides. The first question is 
this: Which 2-form should we integrate over a patchlike 2-segment x to get the 
area of its image? By definition, 

// M = // n(x u , x„) du dv. 
t See remarks and reference following Theorem 8.5 of Chapter VII. 



284 GEOMETRY OF SURFACES IN E 3 [Chap. VI 

Thus we want a 2-form whose value on x„,x„ is 

|| x tt X x, || = VEG - F*. 

In general, a 2-form n such that 

m(v, w) = ± || v X w || for all v, w 

is called an area form. Such a form assigns to every pair of tangent vectors 
v, w either plus or minus the area of the parallelogram with sides v and w. 

7.4 Lemma A surface M has an area form if and only if it is orientable. 
On a (connected) orientable surface M there are exactly two area forms, 
which are negatives of each other. (We denote them by dM and —dM.) 

Proof. If v and w are linearly independent, then || v X w || > 0; thus 
area forms are nonvanishing. Hence, by Definition 7.4 of Chapter IV, a 
nonorientable surface cannot have an area form. 

Now suppose that M is an orientable surface in E 3 . The proof of Theorem 
7.5 of Chapter IV actually establishes a one-to-one correspondence be- 
tween normal vector fields on M and 2-forms on M. If U is a unit normal, 
then the associated 2-form dM is an area form, since 

dM(v, w) = U(p)»y X w = ± || v X w ||. 

(In Fig. 6.18 this number is positive, since v X w points in the same direc- 
tion as U(p), but if v and w were reversed, we would get the negative sign.) 
Thus the two unit normal vector fields on M determine the two area 
forms dM and — dM on M. | 

To orient an orientable surface is to pick one of its two area forms, since 
that amounts to the same thing as picking one of its unit normals. 



V X W 




FIG. 6.18 



Sec. 7] INTEGRATION AND ORIENTATION 285 

Finding area is not a really typical integration problem, since area is 
always positive. Thus to find area by integrating an area form we must be 
careful about signs. Suppose x is a patchlike 2-segment in a surface oriented 
by an area form dM. By definition, 



/ / dM = // dM(x u , x„) du dv. 



Now there are two cases: 

(1) If dM(x u , x„) > 0, we say that x is positively oriented. Then by defi- 
nition of area form, 

dM(x u , x„) = || x u X x„ ||; 

hence // x dM is the area of x(R). 

(2) If dM(xu, x v ) < 0, we say that x is negatively oriented. Then 

dM(x u , x v ) = — || x u X x„ ||, 

hence J/ x dM is minus the area of x(R). 

Thus to find the area of a pavable oriented region by integrating its 
selected area form, we cannot use an arbitrary paving; the paving must be 
positively oriented, that is, consist only of positively oriented patchlike 2- 
segments. Then 



A(<?) = T,M*(Ri)) = Z// 



dM. 



Now we replace the area form by an arbitrary 2-form to get the definition 
we are looking for. 

7.5 Definition Let v be a 2-form defined on a pavable oriented region (P 
in a surface. The integral of v over (P is 






where xi, • • • , x k is a positively oriented paving of (P. 

There is a consistency problem in this definition: We must know that 
two different positively oriented pavings of (P give the same value for the 
sum on the right. A detailed proof would be somewhat long; the general 
scheme is given on page 103 of Hicks [5]. 

Since compact surfaces are pavable, the definition above gives in particu- 
lar the integral of a 2-form over a compact oriented surface. 

7.6 Remark Improper integrals. We have defined area and the integra- 
tion of forms for compact surfaces; however, the notion of area can easily 



286 GEOMETRY OF SURFACES IN E 3 [Chap. VI 

be extended to a noncompact surface N. We define the area of N to be the 
least upper bound of the set of all areas of payable regions (P in N: 

A(N) = l.u.b. A(<P). 

Thus A(N) = +00 if no finite upper bound exists. 

By contrast, it is generally impossible to assign a value — finite or in- 
finite — to the improper integral /JV / dN. The special case / ^ 0, however, 
may be handled in the same way as area; we set 

ff fdN = l.u.b. If fdN ((P pavable in N) 
JJ N JJ(p 

For / ^ 0, switch to greatest lower bound. Thus values + oo and — °o are 
possible in these two cases. Now let 6*1, (P 2 , • • • be a sequence of pavable 
regions in N such that (P* is contained in (P f+1 , and every pavable region in 
N is contained in some G\. It then follows that 

lim ff fdN = ff f dN. 

(We used the corresponding fact for area in (3) of Example 7.2.) 

If (P is a pavable region in a surface M oriented by dM, we have seen 
that ff<p dM is the area of (P. More generally, ffy f dM gives the integral 
of a function / over (P — an obvious analogue of the usual integral J 6 / dx 
from elementary calculus. We turn now to an important geometric applica- 
tion of this idea. 

7.7 Definition Let K be the Gaussian curvature of a surface M, and let 
(P be a pavable region in M oriented by dM. Then the number 



IL 



KdM 



is called the total Gaussian curvature of (P. 

When (P is an entire compact oriented surface M, we get the total Gaus- 
sian curvature of M. The total is an algebraic one: Negative curvature at 
one place may cancel positive curvature at another. 

To compute the total curvature, Definition 7.5 shows that it suffices 
to know how to integrate the 2-form K dM over patchlike 2-segments. But 

jf KdM = ff x*(K dM) 

= ff K(x)x*(dM) = [ f K{xWEG - F 2 du dv 
with the usual notation for x: R -> M. Then K(x) may be computed 



Sec. 7] INTEGRATION AND ORIENTATION 287 

explicitly by Corollary 4.1 of Chapter V or by Lemma 6.3. Luckily the ori- 
entation problems here solve themselves automatically; see Exercise 4(c). 

7.8 Example Total curvature of some surfaces 

(1) Consfanf curvature. If the Gaussian curvature of M is constant, then 
its total curvature is 

ff KdM = K jf dM = K A(M). 

Thus a sphere of radius r has total curvature 4ir (since K A(M) becomes 
(l/r 2 )(4irr 2 )), and the bugle surface has total curvature — 2t (since 
K A (M) becomes ( - 1/c 2 ) (2ttc 2 ) ) . 

(2) Torus. Let x be the 2-segment used on the torus T in Example 7.2. 
By this example the area form dT has the coordinate expression 

x*(dT) = y/EG - F 2 du dv = r(R + r cos u) du dv 

But in Example 6.1 of Chapter V we computed 

K(x) = (cos u)/r(R + r cos u) 

for this same x. Hence, the torus has total curvature 



J J KdT = j j cos u du dv = 0. 



Thus on the torus the negative curvature of its inner half exactly balances 
the positive curvature on its outer half, giving total curvature zero. 

(3) Catenoid. This surface is not compact, and its area is infinite; never- 
theless its total curvature^ — treated by the remark above as an improper 
integral — is finite. On the rectangle R: — a ^ u ^ a, O^vfS 2ir, the 
parametrization in Example 6.1 of Chapter V becomes a patchlike 2- 
segment covering the region between the parallels u = —a and u = a. 
(Fig. 6.19). From Example 6.1 of Chapter V, we get 

K(x) = _1 



c 2 cosh 4 (u/c) 
and 

x*(dM) — y/EGdudv = c cosh 2 (u/c). 
Hence the region has total curvature 

ff K JM — — f f^ ^ U ^ V — —4 t h (^\ 

JJ X J -Jo c cosh 2 (u/c) \c / 



288 



GEOMETRY OF SURFACES IN E 3 



[Chap. VI 




FIG. 6.19 



As a — * oo, this region expands to fill the whole surface; thus the total 
curvature of the catenoid is 



I 



K dM = — 4x lim tanh 



©- 



= — 4t. 



The total curvatures computed above are 4x, — 2ir, 0, and — 4* — a 
rather special set of numbers. Furthermore, none depends on the particu- 
lar "size" (radius r, constant c, • • • ) of its surface. A partial explanation is 
provided by Corollary 7.10; a rather deeper one comes in Chapter VII, 
Section 8. 

If F: M — * iV is a mapping of oriented surfaces, then the Jacobian J of F 
is the real-valued function on M such that 

F*(dN) = J dM. 

We can get an idea of the geometric meaning of J by arguing as in the 
special case at the beginning of this section. If v and w are very small 
tangent vectors at a point p of M, they span a parallelopiped in T P (M) 
which approximates a small region AM in M . The character of the deriva- 
tive map F# is such that F*(v) and F*(w) are the sides of a parallelogram 
in T F ( P) (N) approximating the image region F(AM), as shown in Fig. 
6.20. By the definition of Jacobian we have 



J(p)dM(y,w) = (F*dN)(v,w) = dN(F*v,F*w). 



(*) 



Now || v X w || is approximately the area of AM (and similarly for 
F{AM)). Hence by taking absolute values we get 

| J(p) | (area of AM) ~ area of F{AM). 

Thus | J (p ) | gives the rate at which F is expanding area at p. Furthermore 
if AM is positively oriented, that is, dM(\, w) > 0, then (*) shows that 
the sign of dN(F*\, F*w) is the same as that of J(p). Thus the sign of 



Sec. 7] INTEGRATION AND ORIENTATION 289 



**(w) 







FIG. 6.20 



«/(p) tells whether F preserves or reverses the orientation of AM. 
In this context we call the number 



ff JdM = If F*(dN) 



the algebraic area of F(M). The discussion above shows that, roughly 
speaking, each small region AM in M contributes to this total the algebraic 
area of its image F(AM) : 

(1) Positive, if the orientation of F(AM) agrees with that of N; 

(2) Negative, if these orientations disagree (so F has turned AM over); 

(3 ) Zero, if F collapses AM to a curve or a point. 

Let us consider what this means in the case of the Gauss mapping (Exer- 
cise 4 of Section 1 of Chapter V). 

7.9 Theorem The Gaussian curvature K of an oriented surface M C E 3 
is the Jacobian of its Gauss mapping G: M — *■ 2. 

(Here 2 is the unit sphere, oriented by the outward normal U or the 
corresponding area form d2.) 

Proof. If U = 2 gJJi is the unit normal orienting M, then the cor- 
responding Gauss mapping is G = (g it g 2 , ffr). Notice that if S is the shape 
operator of M given by U, then 

-S(v) = V„£7 = 2vy[/ i (p) 

and by Theorem 7.5 of Chapter I, 

G*(v) =2Y[g i ]U i (G(p)). 

Hence (r*(v) and — S(v) are parallel for any tangent vector v to M, as 
shown in Fig. 6.21. 

To prove the theorem we must show that 

KdM = G*(d2), 

so we evaluate these 2-forms on an arbitrary pair of tangent vectors to M. 
Using Lemma 3.4 of Chapter V, 



290 



GEOMETRY OF SURFACES IN E 3 



[Chap. VI 



U(p) 




U(G(p)) 



G(P) 



FIG. 6.21 



(KdM)(v,w) = K(p) dM(v,w) = K(p) f/(p).v X w 

= U(p)-S(v) X £(w). 

On the other hand, 

(G*d2)(v,w) = d2(G*v, G*w) = U(G(p))-G*v X G*w 

Now a triple scalar product depends only on the Euclidean coordinates of 
its vectors, so G*(\) and Cr*(w) may be replaced by the parallel vectors 
— S(v) and — S(w). Furthermore, by the definition of G and the special 
character of the unit sphere 2, the vectors C/(p) and U(G(p)) are also 
parallel (Fig. 6.21). Thus the two triple scalar products above are the 
same — and the proof is complete. | 

7.10 Corollary The total Gaussian curvature of an oriented surface 
M C E 3 is the algebraic area of the image G(M) of its Gauss mapping 
G:M^2. 

To prove this it suffices to integrate the form 

KdM = G*(d2) 

over M . 

Algebraic area can be tricky when the mapping F: M —> N folds M 
many times over the same regions in N. Thus for practical purposes, the 
following special case of Theorem 7.10 is simpler, since it involves only 
ordinary area. 

7.11 Corollary If (R is an oriented region in M CZ E on which (1) the 
Gauss map G is one-to-one (U is not parallel at different points of (R), and 
(2) either K ^ or K ^ 0, then the total curvature of (R is plus or minus 
the area of G(6i), where the sign is that of K. Furthermore this area does 
not exceed 47r. 

(The proof uses improper integrals.) For example, consider the Gauss 
mapping of an oriented torus. Now G collapses the top and bottom circles 
of T (where K = 0) to the north and south poles of 2. If, as in Fig. 5.21, 
and S are the outer and inner halves of T, then G maps (where K ^ 0) 



Sec. 7] 



INTEGRATION AND ORIENTATION 



291 



in one-to-one fashion onto the whole sphere Z— and does the same for S 
(where K ^ 0). Thus T has total curvature + A (2) - A (2) = 0, as we 
found in Example 7.8 by an explicit integration. 

As another example we find that the entire bugle surface B satisfies the 
hypotheses in Corollary 7.11. In fact, as Fig. 6.22 suggests, the Gauss 
mapping carries its profile curve in one-to-one fashion onto a quarter of a 
great circle in 2. Thus by moving U around the parallels of B, we see that 
G is one-to-one from B onto an open hemisphere (minus its central point, 
since the rim is not part of B). Thus the total curvature of the bugle is 
— (|) A (2) = — 2?r. Furthermore, since the bugle has constant curvature, 
we can find its area without explicit integration: Total curvature — 2?r 
divided by (constant) curvature K = - 1/c gives the area 2ttc 2 , as found 
in Example 7.2. 

On an oriented surface, the ambiguity in the measurement of angles 
mentioned in Chapter II, Section 1, can be reduced. If the surface is ori- 
ented by an unit normal U, then for every tangent vector v to M , U X v 
is a tangent vector orthogonal to v. We shall think of U X v as v rotated 
through +90°. Then if v and w are unit tangent vectors at a point of M, 
a number # is defined to be an oriented angle from v to w provided 

w = cos # v + sin & (U X v). 

(Fig. 6.23). All oriented angles from v to w thus have the form & + 2im, 
where n is an arbitrary integer. (The same scheme applies to any pair of 
nonzero tangent vectors: It suffices to divide by their norms to get unit 
vectors. ) 

Consistency is the essence of orientability— in studying an oriented 
surface oriented by an area form dM we shall always use positively oriented 
patches, dM(x u , x v ) > 0, and positively oriented frame fields, for which 

dM(E u E 2 ) = +1. 




Profile curve 



FIG. 6.22 



292 



GEOMETRY OF SURFACES IN E 3 



[Chap. VI 




FIG. 6.23 

(Note that by definition of area form the only possible values of dM on a 
frame are ±1.) For a positively oriented frame field, we can now give 
geometric meaning to the wedge product of its dual forms : 

dM = 0! a 2 

on the domain of the frame field. To prove this useful fact, it suffices to 
note that both sides have the same value, +1, on E x , E 2 . The second struc- 
tural equation (Corollary 2.3) then becomes da>u = — K dM. 



1. For a Monge patch 



EXERCISES 



x(u,v) = (u,v,f(u,v)), 



show that the area of x(D) is given by the usual formula from ele- 
mentary calculus. Deduce that A(x(D)) ^ A(D). 

2. Find a formula for the area of an arbitrary surface of revolution, and 
interpret it as A = 2irLh, where L is the length of the profile curve and 
h is its average distance from the axis of revolution (Pappus). 

3. Find the area of the following surfaces: 

(a) The region in the saddle surface z = xy covering the disc x 2 + y 2 

^ c in the xy plane. 

(b) Catenoid (c) The Mobius band (Ex. 7 of IV.7). 

4. Let M be a compact surface oriented by dM; let — M be the same 
surface oriented by the other area form —dM. Prove 

(a) / / (civi + C2P2) = c\ I J vi + c 2 / J V2 (ci, c 2 constant). 



Sec. 7] INTEGRATION AND ORIENTATION 293 



<»/L'~/JL 



(Hint: If x(u, v) = x(v, u), show that x and x have opposite orienta- 
tions; then look ahead to Exercise 21.) 

(c) IL fdM = IL M f( ~ dM) - 

(d) UJ^^ythenff fdM£ fj g dM 

(Note the effect of / = or g = 0.) 

Property (c) shows how to define the integral of a, function over a 
compact surface that is merely orienta&Ze; either choice of area form 
leads to the same result. In particular, the total curvature of a com- 
pact orientable surface is now well-defined. 

Total curvature of surfaces of revolution. On a surface of revolution M 
with profile curve a, let Z a & be the region ("zone") between the paral- 
lels through a (a) and a (b). 

(a) Show that the total curvature of Za& is 27r(sin <p a — sin <pb), where 
these are the slope angles of a at a (a) and a (b) — measured rela- 
tive to the axis of revolution (Fig. 6.24). 

(b) Deduce that every surface of revolution whose profile curve is 
closed has total curvature zero. 

If the profile curve a is not closed, then it is one-to-one on some open 
interval A < t < B. Define the total curvature in this case to be 

27r(lim a -A sin <p a — linu-B sin v?&) 

provided both limits exist. 

(c) Test this formula on the bugle and catenoid. 



a'ib) 




FIG. 6.24 



294 GEOMETRY OF SURFACES IN E 3 [Chap. VI 

6. Show that the Gauss mapping of a surface M (Z E is conformal if 
and only if M is part of either a sphere or a minimal surface. 

7. Find the total curvatures of the constant curvature surfaces of revolu- 
tion (Chapter V, Section 6, and Exercises); deduce their areas. 

8. The area forms of E 2 are, as expected, ± du dv (since du and dv are 
the dual forms of a frame field). The natural orientation of E 2 is by du dv; 
this orientation is assumed unless the contrary is explicitly stated. 

(a) Using the general definition in the text, show that the Jacobian 
of a mapping 

F = (/,<?): E 2 -*E 2 

is given by the usual formula 

" = JuQv JvQu- 

(b) Show that the Jacobian of a patch x: D — » M in an oriented sur- 
face (D connected) is ± y/EG — F 2 , where the sign depends on 
whether x is positively or negatively oriented. 

9. Let M be a ruled surface whose rulings are entire straight lines, and 
assume that K < 0. 

(a) Show that the total curvature of M is —2L(8), where 5 is the 
director curve, with || 5 || = 1. 

(b) Compute the total curvature of the saddle surface M : z = xy 
by this method, and check the result using Corollary 7.11. 

10. Total curvature of quadric surfaces 

(a) Find the total curvature of an arbitrary hyperbolic paraboloid, 
elliptic paraboloid, and ellipsoid. 

(b) Show that the total curvature of the hyperboloid of revolution 
M: (x 2 + y)/a - z/c = 1 is -lira/Va 2 + c 2 . 

Thus the total curvature of an elliptic hyperboloid is dependent on 
the particular "dimensions" of the surface — and the same is true for 
an elliptic hyperboloid of two sheets. 

11. A simple region S in M is a region that can be expressed as the image 
F(D) of the disc u 2 + v ^ 1 in E 2 under a one-to-one regular mapping 
F. Show that S can be paved by a single patchlike 2-segment x in 
such a way that for any 1-form we have 



//.*-/. 



<t>, 



where a is an edge curve of x. (Hint: See Ex. 12 of IV.6.) 
12. (Continuation) 



Sec. 7] INTEGRATION AND ORIENTATION 295 

(a) If S is a simple region in E 2 , show that the area of S is 

- / (u dv — v du) , 

where a is the "boundary curve" of S. 

(b) Find the area of the region in E 2 enclosed by the ellipse 

2 / 2 I 2 /7 2 i 

u /a + v /b = 1. 

13. Exercise 7 of Section 8, Chapter VII, will show that if <£ is any 1-form 
on a compact oriented surface, then 



I 



= 0. 



Combine this result with Exercise 3 of Section 2 to show that if h 
is the support function of M C E , then 

A(M) + ff hH dM = ff HdM + // hK dM = 0. 

Check these formulas on a sphere of radius r, oriented by the outward 
unit normal. 

14. Write & = 4- (u, v) if # is an oriented angle from u to v. Show that 

(a) If t? = 4 (u, v) and <p = 2$. (v, w), then & + <p = 2$. (u, w). 

(b) If # = 2$.(u,v), then -# = 2$_(v,u). 

15. Let a: I — » M be a curve in an oriented surface M. If V and W are 
nonzero tangent vector fields on a, show there is a differentiate 
function & on / such that &(t) is an oriented angle from V(t) to JT (<) 
for each tin I. We call # an angle function from V to W. Note that any 
two differ by an integer multiple of 2x. (Hint: Reduce to Ex. 12 of 
ILL) 

16. A mapping F: M — > N is area-preserving provided the area of any 
pavable region (R in M is the same as the area of its image F((R) in N. 
(Note that such a mapping must be one-to-one.) Show that: 

(a) A one-to-one mapping F: M — > N is area-preserving if 

EG - F 2 = EG - F 2 

for every patch x in M, with x = ^(x) in N. (Hint: Show that 
JP carries pavings to pavings.) 

(b) Isometries are area-preserving; isometric surfaces have the same 
area. Include the noncompact case. 

(c) The mapping (1) in Example 5.2 of Chapter IV is area-preserving 
but not an isometry. Deduce the standard formula for the area 



296 GEOMETRY OF SURFACES IN E 3 [Chap. VI 

of a zone in the sphere. (It suffices in (a) to consider a single 
parametrization x if it covers all of M.) 

17. If F: M — > N is a mapping of oriented surfaces with Jacobian J, show 
that: 

(a) F is regular if and only if J is never zero. 

(b) F is area-preserving if F is one-to-one and J = ±1. (The con- 
verse is true also.) 

(c) If F is an isometry, then J = ±1, but the converse is false. 

Since we are dealing only with connected surfaces, (a) shows that all 
such mappings F separate into two classes: orientation-preserving if J > 0, 
orientation-reversing if J < 0. Except for patches (Exercise 8), we use this 
notion mainly in the easy case where F is an isometry. 

18. If F: M — ► M is an isometry of oriented surfaces, show that F is orienta- 
tion-preserving if and only if any one of the following conditions hold: 

(a) F*(dM) = dM. 

(b) F*(U(p) X v) = U(F(p)) XF*(v) for all tangent vectors v to M 
at p (U and U the unit normals orienting M and M ). 

(c) C/(p).vXw = l/(F(p)).F*(v) Xf*(w) for all pairs of tangent 
vectors. 

(d ) For any positively oriented frame field E h E 2 on M , F* (E x ) , F* (E 2 ) 
is a positively oriented frame field on M . 

19. If F: M — > N is an orientation-preserving diffeomorphism of compact 
oriented surfaces, show that 



w IL f ' m - //, * 



for any 2-form on N. (Hint: Ex. 8 of IV.6.) 

(b) Deduce that total curvature is an isometric invariant for compact 
orientable surfaces. 

(c) Extend (b) to the noncompact case, assuming either K ^ or 
K ^ 0. 

20. Gauss mapping G of some minimal surfaces. Prove : 

(a) Catenoid. G is one-to-one and its image covers all the sphere 
except two points. 

(b) Helicoid. The image of G omits exactly two points of the sphere, 
and each point of the image is hit by an infinite number of points 
of the helicoid. 

(c) Scherk's surface (Ex. 21 of V.4). Same as the helicoid except that 
exactly four points are omitted. {Hint: Consider Z = Vg on one 
of the vertical lines.) 

What are the total curvatures of these surfaces? 



Sec. 8] CONGRUENCE OF SURFACES 297 

21. In an oriented surface M let x and y be patchlike 2-segments with 
the same image x(D) = y(E). For any 2-form v show that 



/// - * II; 



with plus sign if x and y have the same orientation (positive or negative), 
minus sign if opposite orientation. (Hint : examine the sign of the Jacobian 
of y _1 x and use the change of variables formula for double integrals.) 



8 Congruence of Surfaces 

Two surfaces M and M are congruent provided there is an isometry F 
of E 3 that carries M exactly onto M. Thus congruent surfaces have the 
same shape — only their positions in E 3 can be different. For example, any 
two spheres of the same radius are congruent (use the translation carrying 
one center to the other), and the surfaces 



M: z = xy and M: z = 



x + y 



are congruent under a 45° rotation about the z axis. 

To simplify the exposition, we shall assume that the surfaces we deal 
with in this section are orientable as well as connected. 

8.1 Theorem If F is a Euclidean isometry such that F(M) = M, then 
the restriction of F to M is an isometry F = F | Af: M — ► M of surfaces. 
Furthermore, if M and M are suitably oriented, then F preserves shape 
operators; that is, 

F*{S(j)) = S(F*(v)) 

for all tangent vectors v to M. 

In short, congruent surfaces are isometric and have essentially the same 
shape operators. We emphasize, however, that isometric surfaces need not 
be congruent, since, as we have seen, they may have quite different shapes 

E3 

Proof. We know from Chapter IV, Section 5, that the restriction 
F: M — > M is a mapping. Furthermore the derivative maps of F and F agree 
on tangent vectors to M. In fact, if v is tangent to M , then v is the initial 
velocity of some curve a. in M — and since F = F | M , we have 

F(a) = F(a). 
Thus 

F*(v) = F(a)'(0) = F(a)'(0) = F*(v). 



298 



GEOMETRY OF SURFACES IN E 3 



[Chap. VI 



It follows immediately that F* preserves dot products of tangent vectors 
to M, for F* has this property for all pairs of tangent vectors (Corollary 
2.2, Chapter III). Also, F: M —*■ M is one-to-one (since F is) and onto 
(by hypothesis); hence F is an isometry of surfaces. 

Finally, we show that F preserves shape operators. If M is oriented by 
the unit normal U, then since F* preserves dot products (and agrees with 
F* on M), it follows that F*(£/) has unit length and is everywhere normal 
to F(M) = M. Thus one of the unit normals on M, say U, has the property 
that 

F*(£/(p) = U(p) where p = F(p). 

If S and S are the shape operators on M and M derived from U and U, 
respectively, we shall show that 

F*(£(v)) = £(F*(v)). 

Again let a be a curve in M with initial velocity v. Thus F(a) is a curve 
in M with initial velocity F*(v). If U is restricted to a, and U is restricted 
toF(a), then F*(t/) = U (Fig. 6.25). Since F* preserves derivatives of 
vector fields, we get 

F*(S(v)) = -F*(t/'(0)) = -U'(0) = £(F*(v)). 
But v and S (v) are tangent to M; hence F* may be replaced by F*. | 

Our goal now is the converse of the preceding theorem, that is: If 
M and M are isometric and have the same shape operators, then M and M 
are congruent. This is the analogue of the basic result (Theorem 5.3 of 
Chapter III) for curves. The condition M and M isometric corresponds to 
the hypothesis that a and /? are unit-speed curves defined on the same 




FIG. 6.25 



Sec. 8] 



CONGRUENCE OF SURFACES 299 



interval, and, of course, "same shape operators" corresponds to 

K = K, T = ±f . 

8.2 Lemma Let F: M -*• M be an isometry of oriented surfaces in E 3 
that preserves shape operators (as in Theorem 8.1 ). Let E u E 2 be a tangent 
frame field on M, with E u E 2 the transferred frame field on M . If E 3 _ and 
E 3 are the unit normals orienting M and M, then E u E 2 , E 3 and E u E 2 , E z 
are adapted frame fields on M and M. For the connection forms of these 
frame fields, we have 

F*{a>ii) = on (1 ^ *J ^ 3). 
Proof. We already know from Lemma 5.3 that F*(wi 2 ) = W12. It remains 
to prove that 

F*(w i3 ) = &>a, fori = 1,2. 

But Corollary 1.5 shows that this merely expresses the preservation of 
shape operators in terms of connection forms. In fact, for j = 1, 2, 

F*(ua){Ei) = 5x»(F*Ej) = S(F*^,-)-^ = F*(S{Ei))'F*{Ei) 

= SEj-Ei = uaiEj). 

Hence the forms F*(wi 3 ) and w t - 3 are equal. I 

8.3 Theorem Let F: M -*• A^ be an isometry of oriented surfaces that 
preserves shape operators; that is, 

*V0S(v)) = S(F*(v)) 

for all tangent vectors v to M. Then M and M are congruent; in fact there 
is an isometry F of E 3 such that F = F | M. 

(If it should happen that 

F* («(▼)) = -8(F*(y)) 
for all tangent vectors, then it suffices to reverse the orientation of either 
M or M to get the hypothesis as stated.) 

Proof. Fix a point p of M, and let E 3 and E z be the unit normals orient- 
ing M and M. By using Corollary 2.3 of Chapter III, it is easy to show 
that there is a unique isometry F of E 3 which agrees with F at the selected 
point p in the sense that 

F(p) = F(p) 
F* (v) = F* (v) for every tangent vector v to M at p 

F*(£ 3 ( P )) = Et(F(p)). 
We shall show that F is the required Euclidean isometry, in other words, 



300 GEOMETRY OF SURFACES IN E 3 [Chap. VI 




FIG. 6.26 



that F(q) = F(q) for an arbitrary point q of M . Thus if a is a curve in M 
from p to q, it suffices to prove that F(a) = F(a). 

There is no loss of generality in assuming that a lies in the domain of an 
adapted frame field E lf E 2 , E 3 on M. (If not, we could break a up into 
segments for which this is true, and repeat essentially the following proof 
for each segment in turn.) Our plan is to use the general criterion Theorem 
5.7 of Chapter III to show that the curves F(a) and F(a) are identical. 

A. The curve F(a). Restrict the frame field E u E 2 , E 3 to the curve a 
(Fig. 6.26). Then by the connection equations, 

El = V a ,Ei = 2 aaicDEj (1 ^ i ^ 3). 

Now apply F* to this equation; since F* is linear and preserves derivatives, 
we get 

(F*E<)' = Z «*,(«') F* (Ej) (1 £ » £ 3). (Al) 

iucts; hence F*E lf F*E 2 , F*E 3 is the frame field 
Furthermore, 

F(a)'.F*^- = d'Ei (A2) 



Also F* preserves dot products; hence F*E lt F*E 2 , F*E 3 is the frame field 
on the image curve F(a). Furthermore, 



since 

V{a)' = F*(«'). 

B. The curve F (a) = a. Use the isometry F to transfer the tangent 
frame field E u E 2 to a tangent frame field E u E 2 on M. With the unit nor- 
mal vector field E z , we now have an adapted frame field E lf E 2 , E s 
on M. We restrict it to the image curve F(a) = a, and use the connection 
equations as above to get 

El = Z Z>ij(a)Ej (1 ^ i ^ 3). (Bl) 

Furthermore, we assert that 

ol'Ei = cl*Ei (1 ^ i ^ 3). (B2) 

For i = 1, 2 this follows immediately from the definition of Ei, E 2 , since 



Sec. 8] CONGRUENCE OF SURFACES 301 

F* is an isometry and 

ol = F(a)' = F*(a). 

For i = 3, both sides are zero, since a and 5 are curves in M and M , re- 
spectively. 

C. Comparison of F(a) and F (a) = a. The construction above and the 
assumption that F preserves shape operators has exactly reproduced the 
hypotheses of the preceding lemma; hence F*(uij) — ua for 1 ^ i,j ^ 3. 
Thus 

««(»') = S><j(F*(d)) = (F*«<y)(«) = mi(d). 

Using this fact we deduce from (Al) and (Bl) that 

(F+EiY'F+Ej = Ei'-E; (1 ^ ij ^ 3). (CI) 

Comparing (A2) and (B2), we get 

F(a)''F*Ei = a'-Ei (1 ^ t ^ 3). (C2) 

And by our initial construction we have 

F*^i = Ei at the point p = a(0) 

= F(a(0)) (1 ^ t ^ 3). 

Refering to equations (J) in Theorem 5.7 of Chapter III, we observe 
that the three equations (CI), (C2), (C3) are precisely what is needed 
to conclude that F(a) = a; that is, 

F(a) = F(a). I 

This theorem gives a formal proof that the shape operators of a surface 
M in E 3 do, in fact, completely describe its shape. 



EXERCISES 

1. A surface M C E 3 is rigid provided every surface isometric to M is 
congruent to M . Deduce from Liebmann's theorem that spheres are 
rigid. 

2. If a, /8: / — * E 3 are unit-speed curves with K a = «/s > and t„ = rp, 
show that their tangent surfaces are congruent. (Compare Ex. 5 of 
VI.4.) 

3. If M and N are congruent surfaces in E 3 , and F is a Euclidean isometry 
such that F(M) = N, prove that F | M preserves Gaussian and mean 



(C3) 



302 GEOMETRY OF SURFACES IN E 3 [Chap. VI 

curvature, principal curvatures, principal directions, umbilics, asymp- 
totic and principal curves, and geodesies. Which of these are preserved 
by arbitrary isometries F: M —*■ N? (Hint: Orient locally, and ignore 
ambiguity of signs for H, k ly and A; 2 .) 

4. If F: 2 — ► 2 is an isometry of spheres, show that there is a Euclidean 
isometry F such that F = F | 2. 

5. Let M be the saddle surface (z = xy). A rotation of 90° followed by 
a reflection in the xy plane yields an orthogonal transformation C of 
E whose matrix is 

/0 -1 0\ 



\0 -1, 

(With our conventions, the columns of the matrix are C(ui), C(ii2), 
C(us), where u» is the ith. unit point.) 

(a) Prove that C(M) = M. 

(b) Let F = C | M: M -> M. Orient M (as domain of F) by the unit 
normal U such that £7(0) = u 3 . For which orientation of M (as 
image of F ) does F preserve shape operators? 

6. In the general description of a surface of revolution on page 129 (M 
obtained by revolving C around A), let A be the line through p in 
the direction of a unit vector ei, and let 

a{u) = p + g(u) e x + h(u) e 2 

be a parametrization of C, where e 2 is a unit vector orthogonal to d. 

(a) Find a regular mapping x: D — » E whose image is the set M. 
Then prove: 

(b) M is congruent to a surface of revolution in the special position 
given in Example 2.5 of Chapter IV. 

(c) M is a surface in E . 

(d) Two surfaces of revolution are congruent if and only if they can 
be described in this way by the same pair of functions g, h. 

7. If M is a surface in E 3 , a Euclidean isometry F such that ¥(M) = M 
is called a Euclidean symmetry of M. Show that 

(a) The set of all Euclidean symmetries of M forms a subgroup §>(M) 
of the group S of all isometries of E 3 (Ex. 7 of III.l). §>(M) is 
called the Euclidean symmetry group of M. 

(b) The Euclidean symmetry groups of congruent surfaces are iso- 
morphic. 

8. Show that the Euclidean symmetry group of any sphere is isomorphic 
to the group of all 3 X 3 orthogonal matrices. 



Sec. 9] SUMMARY 303 

9. Find all eight Euclidean symmetries of the saddle surface M: z = xy. 
Show that they are orthogonal transformations and give their matrices. 

10. Find all Euclidean symmetries of the ellipsoid x 2 /a + y 2 /b 2 + z /c =1, 
where a > b > c. (Hint: Use the fact that Gaussian curvature is 
preserved.) 
11. If M is Scherk's surface (Ex. 21 of V.4) and D is the open square 
— 71-/2 ^ u, v S t/2, show that: 

(a) The image x(Z>) of x (Ex. 4 of V.4) lies in M. 

(b) The portion of M over any open square (Ex. 21 of V.4) is con- 
gruent to x(D). 

(c) The curvature formulas given in the abovementioned exercises 
are consistent. 



9 Summary 

The geometrical study of a surface M in E 3 separates into three distinct 
categories : 

(1 ) The intrinsic geometry of M . 

(2) The shape of M in E 3 . 

(3 ) The Euclidean geometry of E 3 . 

We saw in Chapters II and III that the geometry of E 3 is based on the dot 
product and consists of all concepts preserved by isometriesof E 3 . Similarly, 
we have now found that the intrinsic geometry of M is based on the dot 
product— applied only to vectors tangent to M— and that it consists of 
all concepts preserved by isometries of M . 

The shape of M in E 3 is, in a sense, a link between these two geometries. 
For example, Gaussian curvature K is an essential feature of the intrinsic 
geometry of M, and the shape operator S dominates category (2) — thus 
the equation 

K = det S 



shows that the geometries (1) and (3) can be harmonized only by means 
of restrictions on (2). Stated bluntly: Only certain shapes are possible in 
E 3 for a surface M with prescribed Gaussian curvature. A strong result of 
this character is Liebmann's theorem, which asserts that a compact surface 
in E 3 with K constant has only one possible shape — spherical. 

In the last two chapters, the computation of explicit examples has been 
mostly in terms of coordinate patches (Gauss), but the theory itself has 
been expressed in terms of frame fields and forms (Cartan). Historically, 
coordinates were used for the theory as well, but by now the Cartan ap- 
proach has largely won the day. We have seen in Section 6 that the two 
approaches are not so far apart when the coordinate patch is orthogonal. 



CHAPTER 



VII 



Riemannian Geometry 



In studying the geometry of a surface in E 3 we found that some of its 
most important geometric properties belong to the surface itself and not 
the surrounding Euclidean space. Gaussian curvature is a prime example; 
although denned in terms of shape operators, it belongs to this intrinsic 
geometry, since it passes the test of isometric invariance. As this situation 
gradually became clear to the mathematicians of the 19th century, Rie- 
mann drew the correct conclusion: There must exist a geometrical theory 
of surfaces completely independent of E 3 , a geometry built from the start 
solely of isometric invariants. In this chapter we shall give an outline 
of the resulting theory, concentrating on its dominant features: Gaussian 
curvature and geodesies. Our constant guides will be the two special cases 
which led to its discovery: the intrinsic geometry of surfaces in E 3 , and 
Euclidean geometry — particularly that of the plane E 2 . 



1 Geometric Surfaces 

The evidence from earlier work on the intrinsic geometry of surfaces in 
E (and on Euclidean geometry as well) suggests that we will need the dot 
product on tangent vectors to do geometry on a surface. 

\But to free ourselves of confinement to E 3 , we must begin with an ab- 
stract surf ace M (Chapter IV, Section 8). Since M need not be in E 3 there 
is no dot product — and hence no geometry. The dot product, however, is 
but one instance of the general notion of inner product, and Riemann's 
idea was to replace the dot product by a quite arbitrary inner product on each 
tangent plane of M. 

304 



Sec. 1] GEOMETRIC SURFACES 305 

1.1 Definition An inner product on a vector space V is a function which 
assigns to each pair of vectors v, w in V a number vowso that these rules 
hold: 

(1) Bilinearity: 

(fllVi + O2V2) o W = ttiVi o W + 02V2 o W 
V o (61 Wi + 62W2) = 61V 0W1 + 62V o W2. 

(2) Symmetry: vow = wov. 

(3) Positive definiteness: 

vov^ 0; and v o v = if and only if v = 0. 
On the vector space E 2 the dot product 

VW = V1W1 + V 2 W 2 

is, of course, an inner product, but there are infinitely many others, such 
as, for example, v o w = 2viWi + 3v 2 W2. (See Exercise 8 of Section 2.) 

Shifting then from surfaces in E 3 to abstract surfaces and, from the dot 
product to arbitrary inner products, we get the following definition. 

1.2 Definition A geometric surface is an abstract surface M furnished 
with an inner product, °, on each of its tangent planes. This inner product 
is required to be differentiable in the sense that if V and W are (differentia- 
ble) vector fields on M, then V° W is a differentiable real- valued function 
on M. 

We emphasize that each tangent plane T P (M) of M has its own inner 
product, and they are unrelated save for the differentiability condition — 
an obvious necessity for a theory founded on the calculus. In this definition 
V°W has its usual pointwise meaning: It is the function on M whose 
value at each point p is the number 7(p) ° W(p). An assignment of inner 
products to tangent planes as in Definition 1 .2 is called a geometric structure 
(or metric tensor or "ds ") on M. 

In short : 

Surface + geometric structure = geometric surface 

and we emphasize that the same surface furnished with two different geo- 
metric structures gives rise to two different geometric surfaces. 

1.3 Example Some geometric surfaces. 

(1) The plane E 2 , furnished with the usual dot product on tangent 
vectors, is the best-known geometric surface. Its geometry is two-dimen- 
sional Euclidean geometry. 

(2) A simple way to get new geometric structures is to distort old ones. 



306 RIEMANNIAN GEOMETRY [Chap. VII 

For example, if g > is any differentiable function on the plane, and • is 
the usual dot product, define 



Vo W 



9 2 (p) 

for tangent vectors v and w to E 2 at p. This is a new geometric structure 
on the plane, said to be conformal to that of the dot product (Exercise 1). 
We shall see that (unless g is special) the resulting geometric surface has 
properties quite different from the Euclidean plane (1 ) . 

(3) If ¥ is a surface in E , then the dot product from E 3 applied to 
tangent vectors on M furnishes an inner product making M a geometric 
surface. This, of course, is just what was done in Chapters V and VI. Unless 
some other inner product is explicitly mentioned, it is always assumed that 
a surface in E is made geometric in this way. 

Here a word about terminology is in order. Euclid's name carries geo- 
metric implications. Hence in Chapter I, where geometry did not appear, 
we should have called E 2 the Cartesian plane, reserving the term Euclidean 
plane for the geometric surface (1 ) above. 

From the simple beginning in Definition 1.2, it is rather surprising what a 
rich geometric theory can be built. But as mentioned earlier, examples 
(1) and (3) certainly indicate that the theory is there to be explored, 
and their common features even suggest the kind of results we may expect 
to find. 

The definitions in Chapter VI that are clearly intrinsic in character 
will be used here without further discussion. In particular, an isometry F: 
M — > M of arbitrary geometric surfaces is exactly as denned in Definition 
4.2 of Chapter VI and the geometry of M consists by definition of its iso- 
metric invariants. A frame field on an arbitrary geometric surface M con- 
sists, as usual, of two orthogonal unit vector fields E\, Ei defined on some 
open set of M. The orthonormality equations 

EioEj = Sn (1 ^ i,j ^ 2) 

are expressed, of course, in terms of the inner product of M . As before, 
we derive the dual 1-forms 0i, 02, characterized by di(E,) — b ijy and then 
the connection form un = — «2i, characterized by the first structural 
equations, 

ddi = C0]2 A 02, ddz = G>21 A 01- 

We emphasize once more that these forms 0i, 02, un a-re not invariantly 
attached to the geometric surface M; a different choice of frame field 
Ei, Ei will produce different forms 0i, 2 , wi 2 . Before going any further, we 
had better see how two such sets of forms are related. 



Sec. 1] 



GEOMETRIC SURFACES 



307 



E t 




E, 



*~E t 



\ 

(A) 

FIG. 7.1 



On a small enough neighborhood of a point p, 
careful use of the inverse function cos - (or sin ) E t 
will yield a differentiate function <p such that \^ 

Ei = cos (p Ei + sin <p E 2 . 

We call <p an angle function from Ei, E 2 to Ei, E 2 . 
As shown in Fig. 7.1, there are now two possibili- 
ties for E 2 . Either 

Ei = — sin <p Ei + cos <p E 2 

in which case we say that Ei, E 2 and E lf E 2 have the 
same orientation, or 

E 2 = sin <p Ei — cos <p E 2 , 

which is opposite orientation. 

1.4 Lemma Let E\, E 2 and E u E 2 be frame fields on the same region in 
M. If these frame fields have 

(1) The same orientation, then 

W12 = cow + d(p, and 0i a s 2 = 0i a 6 2 . 

(2) Opposite orientation, then 

«i2 = — (o>i2 + rf^), and 0i a 2 = — 1 a 2 . 

Proof. We discuss only the first case, since the second is obtained from 
it by merely changing signs. By the basis formulas (Lemma 2.1 of Chapter 
VI), the equations 

Ei = cos <p Ei + sin <p E 2 , E 2 = —sin <p E x + cos ^ E 2 
yield 

0i = cos <p 0i — sin £> 2 , 02 = sin <e> 0~i + cos ^02. ( ) 

Applying the exterior derivative to the first of these, we get 

rf0i = — sin <p d<p a 0i + cos <p ddi — cos <p d<p A d 2 — sin <p dd 2 . 
Now we substitute the first structural equations for ddi, dd 2 to obtain 
ddi = (wi2 — d(p) a (sin <p 0i -f cos ^> 2 ) 
= (<ii2 — d<p) a # 2 . 
In the same way we get 

dd 2 = — (a>i2 — dip) a 0j. 

Because the form a>i2 = — co 2 i uniquely satisfies the first structural equa- 
tions, we conclude from the last two equations that coi2 = 0012 — dtp, as 



308 



RIEMANNIAN GEOMETRY 



[Chap. VII 




FIG. 7.2 

required. Direct computation of 0i a 2 using (*) above shows that this 
2-form equals 0~i a 2 . | 

The concept of geometric surface can be used to fill a gap in our earlier 
work. Occasionally we met regular mappings x: D —> E 3 which were not 
parametrizations of any surface in E 3 . For example, the image x(Z>) of D 
might fold back through itself as indicated in Fig. 7.2, so that the definition 
of surface in E fails along the crossing line L. 

This technical difficulty can be eliminated by assigning D (which is a 
surface) not the usual dot product, but instead the induced inner product 

vow= x*(v)«x*(w). 

Thus D becomes a geometric surface, and if x(D) were a surface in E , x 
would evidently be an isometry. In short, D has exactly the intrinsic geome- 
try we might intuitively expect x(D) to have. 

In this chapter, as earlier, the restriction to low dimensions is not essen- 
tial. A surface is the two-dimensional case of the general notion of mani- 
fold (Chapter IV, Section 8). A manifold M of arbitrary dimension fur- 
nished with a (differentiable) inner product on each of its tangent spaces 
is called a Riemannian manifold, and the resulting geometry is Riemannian 
geometry. (Euclidean geometry, as discussed in Chapter III, is the special 
case of Riemannian geometry obtained on the Euclidean space E™, with 
its usual dot product.) A geometric surface is thus the same thing as a two- 
dimensional Riemannian manifold, and the subject of this chapter is two- 
dimensional Riemannian geometry, f 



EXERCISES 

1 . For a conformal geometric structure on the plane (Example 1 .3 ) , show 
that 

tWe would prefer to call a geometric surface a Riemann(ian) surface but this 
term has a firmly established and distinctly different meaning. 



Sec. 1] GEOMETRIC SURFACES 309 

(a) The formula 

vow = || v || || w || cost? (where || v || = -y/v°v) 

gives the same value for the angle ^ # ^ t between v and w as 
in the Euclidean plane E 2 . 

(b) The speed of a curve a = (an, a 2 ) is -y/a'i 2 + a 2 2 /g(a). 

(c) gUi, gU 2 is a frame field with dual forms du/g, dv/g. 

(d) The area forms are ± dudvl g 2 . 

Note that g = 1 gives the usual Euclidean structure. 

2. The Poineari half-plane is the upper half-plane v > furnished with 
the inner product (o) obtained by dividing the dot product at each 
point p by the square of the distance v(p) = p 2 to the u axis: 

vow = v • w/y (p ) . 

For the curve a{t) = (r cos t, r sin t), < t < w, find the speed and 
arc-length function (measure from the top of the semicircle, t = x/2. ) 

3. (a) On a geometric surface M, let V and W be vector fields that are 
linearly independent. Find a frame field for which Ei = V/\\ V ||. 

(b) Deduce an explicit formula for a frame field on the image of an 
arbitrary patch xinM. 

4. If dM is an area form on M and v is an arbitrary 2-form, show that 
there is a function / such that v = f dM. Deduce that a (connected) 
orientable geometric surface has exactly two area forms, ± dM. 

5. Let M be a geometric surface oriented by area form dM. Prove 

(a) On each tangent space to M there exists a unique "rotation by 
+90°," that is, a linear operator J: T P (M) -* T P (M) such that 

ll^(v)|| = ||v||, J(v)° v = 0, 

and 

dM(v,J(y)) > (if v ^ 0). 

(Hint: If E u E 2 is a positively oriented frame field, 

J(E,) = E 2 , J(E 2 ) = -El) 

We call these operators — collectively, for all points of M — the 
rotation operator^ of M . 

(b) J is differentiate (J(V)° J(W) is differentiate for any vector 
fields V, W) and skew-symmetric 

J{V)°W + V°J(W) = 0, 

and J 2 = —I (J applied twice is minus the identity operator). 

t Compare the special case in Exercise 14 of Section 4, Chapter II. 



310 RIEMANNIAN GEOMETRY [Chap. VII 

(c) If M is oriented instead by —dM, then its rotation operator is —J. 

(d) If M is a surface in E and its orientation is given by the unit 
normal U, then J(V) = U X V. 

The operator J serves as a kind of replacement for the unit normal 
on surfaces that are not in E 3 . In particular, (d) shows how the scheme 
given in Chapter VI (page 291) for measuring oriented angles applies 
now to an arbitrary (oriented) geometric surface. 

6. If F: M — > N is a regular mapping of oriented geometric surfaces, show 
that the following are equivalent: 

(a) F is orientation-preserving and conformal (Definition 4.7 of 
Chapter VI). 

(b) F preserves the rotation operators of M and N; that is, 

F*(J(v)) =J(F,(v)) 

for all tangent vectors v to M . 

(c) F preserves oriented angles; that is, if # is an angle from v to w, 
then # is also an angle from F* (v) to F* (w). 

7. (a) Prove that a regular mapping F = (/, g): E 2 — * E 2 is orientation- 

preserving and conformalf if and only if f u = g v , f v = —g u . 

If E 2 is considered as the complex plane, with z = u + iv = (u,v), 
these two equations (the Cauchy-Riemann equations) are necessary and 
sufficient for F to be a complex analytic function z —> F(z). 

(b) Given such a complex function F, show that its scale factor X(z) is 
the magnitude of the (complex) derivative dF/dz. 

8. If the origin is deleted from E 2 , show that the mapping F in (2) of Ex- 
ample 7.3 of Chapter I is orientation-preserving and conformal. What 
is the complex function in this case? 

9. Let D and E be regions in the plane, furnished with conformal geometric 
structures given by functions g\ and gi , respectively. Let D' and E' 
be the same regions, with the usual Euclidean structure. If F: D' — » E' 
is a conformal mapping with scale factor X, prove that F: D — > E is 
conformal with scale factor \gi/gi(F). 



2 Gaussian Curvature 

For arbitrary geometric surfaces, we need a new definition of Gaussian 
curvature. The definition K = det S for surfaces in E is meaningless now, 

t The term conformal is often understood to include preservation of orientation. 



Sec. 2] GAUSSIAN CURVATURE 311 

since it is based on shape operators. But this original definition made K 
an isometric invariant, so it is reasonable to look to the proof of the theo- 
rems egregium (specifically to Corollary 2.3 of Chapter VI) to find a satis- 
factory generalization. 

2.1 Theorem On a geometric surface M there is a unique real- valued 
function K such that for any frame field on M the second structural equation 

dun = —Kdi A 2 
holds. K is called the Gaussian curvature of M. 

Proof. For each frame field E u E 2 , there is (by the basis formulas of 
Lemma 2.1 of Chapter VI), a unique function K such that 

d<t3i2 = —Kdi a 2 . 

But another frame field E u E 2 might a priori have a different function 
K such that 

du l2 = -Kdi A 02- 

What we have to show is consistency: Where the domains of these frame 
fields overlap, there K = K. Since such domains cover all of M (Exercise 
3 of Section 1), we will then have a single function K on M with the re- 
quired property. This consistency will follow immediately from Lemma 1.4. 
First consider the case in which the frame fields have the same orienta- 
tion, so u n = a>i2 + d<p. Hence dun = dun, because d = 0. But then 



Since 



we conclude that 



Kdi A 62 = Kd\ A 02. 

01 A 2 = di A 2 y£ 0, 

K = K. 



When the orientations are opposite, we get dun = —dun but still find that 
K = K, since 

01 A 2 = -0 X A 2 . | 

As noted above, Corollary 2.3 of Chapter VI shows that this general 
definition of Gaussian curvature agrees with the definition K = det S when 
M is a surface in E 3 . The proof of isometric invariance obtained there is 
entirely intrinsic in character, and it thus holds for arbitrary geometric 
surfaces. 

Gaussian curvature is the central property of a geometric surface M ; 



312 RIEMANNIAN GEOMETRY [Chap. VII 

it influences — often decisively — many of the most important properties 
of M. In Section 6 we shall examine the influence of curvature on geodesies 
and in Section 8 its influence on the topology of M . 

To summarize: Geometrical investigations in terms of a frame field 
Ei, E 2 are dominated by its structural equations: 

iddi = a>i2 a Bi, 

\ddi = C021 a S\, 

dun = — Kdi a 2 . 
The first structural equations actually define the connection form 

0>12 = — C021 

of that frame field, while the second structural equation defines the Gaussian 
curvature K of the geometric surface (independent of choice of frame field ) . 
It is already clear from Chapter VI, Section 6, how oi 12 and K may be ex- 
plicitly computed from these implicit definitions. 

2.2 Example Gaussian curvature. 

(1) The Euclidean plane E . If we use the natural frame field Ui, U 2 , 
then the dual 1-forms are merely d x = du, 2 = dv. Since dd\ = dd 2 = 0, 
the identically zero form o>i 2 = satisfies the first structural equations and 
hence is the connection form of Ui, Ui. But then dun = 0, so K = 0. 
The Euclidean plane is flat. This can be no surprise, since E 2 is isometric 
to a plane in E , for which we know that K = 0, since its shape operators 
all vanish. 

(2) The plane with conformal inner product 

v • w 

V o w = 

<7(p) 2 

(See (2) of Example 1.3.) 

The natural Euclidean frame field U\, Ui is no longer a frame field rela- 
tive to this new inner product. U\ and Ui are still orthogonal, but 

UtoTJi = U 2 oU 2 = -,. 
9 2 

Thus gU\, gUi is a frame field. It follows easily that its dual 1-forms are 

a - du a - dv 

V\ — , C/ 2 • 

g 9 

To find the connection form W12, we first differentiate 9 X and Q%. 



Sec. 2] GAUSSIAN CURVATURE 313 

ddi = d ( - J A du 



dd 2 = d ( - J a dv. 

Now d (l/g) = —dg/g 2 and dg = g u du + £» efo. Because dw dw = dvdv = 0, 
we find 

ddi = ( 9 ~ du) a 2 

/- \ () 

dd 2 = -l-^dvj a Bx. 

Comparison with the first structural equations then yields 

wis = - (ft du — g u dv) 
9 

because by (*), this form coi 2 satisfies the first structural equations; by 
uniqueness it must be the connection form. 

To get the curvature we differentiate once more. 

duu = d I - ) a (g v du — g u dv) + - (g™ dv du — g uu du dv) . 
W 9 

From the above we know that 



d (- ) = — r- (g u du + g, dv) 
\9/ 9 2 



and that du = gdi, dv = gd 2 . A simple computation using these facts then 
gives 

dwi2 = (g u 2 + gv — g(guu + g™)) fa A ^a- 
Thus by the second structural equation we conclude that 

K = g(g uu + gw) - (gu + gv). 

The induced inner product discussed on page 308 may be applied in 
other situations. For example, suppose that F: M — > N is a diffeomorphism 
of surfaces (Chapter IV, Section 5) and that N is a geometric surface. 
Then the induced inner product 

vow = F* (v) oF* (w) 

on tangent vectors to M , makes M a geometric surface — and F an isome- 
try. M might be called a "new model" of N; however different it may look, 
it is geometrically identical with AT. 



314 



RIEMANNIAN GEOMETRY 



[Chap. VII 




FIG. 7.3 



2.3 Example 

(1) The stereographic sphere. We have proved in Example 5.5 of Chapter 
IV that stereographic projection P is a diffeomorphism of the punctured 
sphere 2 onto the Euclidean plane E 2 . Now consider S merely as a surface, 
while E is geometric, with its usual dot product. Thus the induced inner 
product makes S a geometric surface which is isometric to E 2 and hence 
flat. If 2 appears round, it is only because we look at it with Euclidean 
eyes; that is, we erroneously assume it must have the dot product of E 3 
as in Chapter V. 

(2) The stereographic plane. Now let us reverse the process in (1 ) . Con- 
sider 2 with its usual geometric structure as a surface in E 3 , and let E 2 be 
merely a surface. 

The inverse P _1 : E 2 — > S of stereographic projection is also a diffeo- 
morphism. The inner product (°) induced by P _1 on E 2 makes E 2 a geo- 
metric surface (the stereographic plane) isometric to S , and thus having a 
curvature K = +1. 

We examine this new stereographic plane more closely. Throughout 
a dot (• ) will as usual denote the dot product, whether of E 2 or E 3 . 

If v and w are tangent vectors to E 2 at q = P(p), let v and w be the 
unique tangent vectors to 2 at p such that P* (v ) = v, P* (w) = w (Fig. 
7.3). Now by Exercise 14 of Section 4, Chapter VI, we know that 



vw = P*(v) • P*(w) 



+ w 



But (P *)* carries v and w back to v and w, so for the induced inner 
product of E 2 we find 



vo W = (P X )*(v) • (P 1 )*(w) = v«w 



- + ¥r 



It follows immediately that this inner product is of the conformal type 
discussed in Example 1.3 with 



1 + 



2 I I 

u + v 



Sec. 2] GAUSSIAN CURVATURE 315 




To visualize this unusual "plane," we may imagine that rulers get longer 
as they move farther away from the origin. Since P is now an isometry, the 
intrinsic distance from p to q (in Fig. 7.4) is exactly the same as the dis- 
tance from p* to q* . Also circles u 2 + v 2 = r 2 , for which r is very large, 
actually have very small stereographic arc length since they correspond 
(under the isometry P) to small circles about the north pole in S . 

2.4 Example The hyperbolic plane. Let us experiment with a change 
of sign in the stereographic inner product above, setting 

2 , 2 
! U + V 

Since g > is necessary, this hyperbolic inner product vow = (l/g )v»w 
is used only on the disc u 2 + v 2 < 4 of radius 2 in the plane. The resulting 
geometric surface is called the hyperbolic plane H. 

In this case 

_ ~ u _ ~~ v a — — — * 

9u n~ , gv — ~n~ ■> ancl g U u — ff™ — — tt > 

so the general computations in Example 2.2 show that 

wi2 = — (u dv — v du) 

and that the hyperbolic plane has constant Gaussian curvature K = —1. 

As a point (u, v) approaches the rim of H, that is, the circle u 2 + v =4 
(not a pait of HI), g(u, v) approaches zero. Thus in the language used 
above, rulers must shrink as they approach the rim, so that H is a good 
deal bigger than one's Euclidean intuition may suggest. For example, 
for # constant let us compute the arc -length function s(t) of the Euclidean 
line segment 

a(t) = (t cos t», t sin 0), S t < 2, 

which runs from the origin almost to the rim. Now a = (cos t?, sin#), 
so a °a' = \/g(a) 2 . But 



316 RIEMANNIAN GEOMETRY [Chap. VII 

g(a(t)) = 1-|, 



so a has hyperbolic speed 

II «'(«) II = 



1 



g(a) (1 - t 2 /4) 



Thus 



s(() = rr^k = 2tanh "1 = ^ 



2 + < 
f 2/4 - " ""— 2 ~ iWfe 2 - « * 

Thus as t approaches 2, arc length s(t) from the origin a(0) to «(£) 
approaches infinity. This "short" segment a actually has infinite hyperbolic 
length. Further properties of the hyperbolic plane will be developed as we 
go along. We shall see that it — and not the bugle surface (Example 6.6 of 
Chapter V) — is the true analogue of the sphere for constant negative 
curvature. 

2.5 Example A flat torus. Let T be a torus of revolution considered 
merely as a surface, and let x: E 2 — » T be its usual parametrization (Exam- 
ple 2.6 of Chapter IV). Now we give T a geometric structure by defining 

It is easy to check that this defines — without ambiguity — an inner product 
on each tangent plane of T. 

Because x*(?7i) = x„ and x*(Uz) = x„, it follows immediately that 
x is a local isometry of the Euclidean plane E 2 onto the geometric surface 
T. Since local isometries also preserve Gaussian curvature, T is flat . Thus 
its geometric structure is different from the usual torus in E , which has 
variable curvature. 

Because this torus T is compact and flat, Theorem 3.5 of Chapter VI 
shows that it can never be found in E . Explicitly, there exists no surface 
M in E 3 that is isometric to T, for then M would also be compact and flat — 
but this is forbidden by the theorem. Here we have a proof that the class 
of geometric surfaces is richer than that of surfaces in E . In the course of 
this chapter we hope to convince the reader that geometric surfaces are 
the natural objects to study and that surfaces in E — however intuitive 
they may seem at first glance — are no more than an interesting special case. 

One should not conclude from the example above that every surface 
can be given a flat geometric structure. Topological subtleties are involved, 
as we shall see in Section 8. 

2.6 Remark So far we have reserved the dot notation (•) for the dot 
product of Euclidean space, and used a small circle (°) to emphasize the 



Sec. 2] GAUSSIAN CURVATURE 317 

generality of the inner product of an arbitrary geometric surface. From 
now on we shall use a dot for all inner products, reverting to the former 
convention only when, as in Example 2.4, the two appear in the same 
context. 



EXERCISES 

1. Derive the dual forms and connection form «i 2 = du/v for the frame 
field vUi, vU 2 on the Poincar6 half-plane (Ex. 2 of VII.l) and show 
that this surface has constant negative curvature K = — 1. 

2. For the conformal geometric structure on the entire plane with 

g = cosh (uv), 

compute the dual forms and connection form of the frame field gU u 
gUi, and derive the Gaussian curvature K. 

3. Find the area A of the disc u 2 -f v S r 2 in the hyperbolic plane. (Hint- 
Find E, F, G for a 2-segment 

x(w,v) = (u cos v, u sin v).) 

What is the area of the entire hyperbolic plane? 

4. The hyperbolic plane of pseudo-radius r is obtained by altering the 
function g in Example 2.4 to g = 1 + (u + v)/Ar 2 . Find its Gaussian 
curvature. 

5. Find the area of the flat torus in Example 2.5. Modify the definition 
so as to produce a flat torus with arbitrary area A > 0. 

6. Show that there is a geometric structure on the projective plane such 
that the natural mapping P: S — * S is a local isometry. Prove that this 
geometric surface 2 cannot be found in E 3 . (The same results hold when 
S is a sphere of radius r and S becomes the projective plane of radius r.) 

7. Show that the plane, furnished with the conformal geometric structure 
such that g = sech u, is isometric to a helicoid. 

8. The identity map \(u,v) = (u,v) of E 2 is a patch with x„ = Ui,x v = U-i. 
Thus if (° ) is a geometric structure on the plane, this patch has 

E = Ui o U u F = U 1 o U 2 , G = Ui o Vi. 

(a) Given any differentiate functions E, F, and G on the plane, such 
that E > 0,G> 0, EG — F 2 > 0, show that there is a geometric 
structure on the plane corresponding, as above, to these functions. 

(b) Show that the method used in Example 2.2 is a special case of 



318 RIEMANNIAN GEOMETRY [Chap. VII 

that of Chapter VI, Section 6, and derive the formula for K in 
Example 2.2 from Lemma 6.3 of Chapter VI. 
{Hint: For (a) define 

vow = EviWi + F(viW 2 + v 2 Wi) + Gv 2 w 2 .) 

3 Covariant Derivative 

The covariant derivative V of E 3 (Chapter II, Section 5) is an essential 
part of Euclidean geometry. We used it, for example, to define the shape 
operator of a surface in E , and in modified form (Chapter II, Section 2) 
to define the acceleration of a curve in E 3 . In this section we will show that 
each geometric surface has its own notion of covariant derivative. 

As in Euclidean space, a covariant derivative V on a geometric surface M 
assigns to each pair of vector fields V, W on M a new vector field V r W, 
and we must certainly require that it have the usual linear and Leibnizian 
properties (Corollary 5.4 of Chapter II). Intuitively, the value of V Y W 
at a point p will be the rate of change of W in the V (p) direction. Thus if 
the connection form uu of a frame field E\, E 2 is to have its usual geometri- 
cal meaning (measuring rates at which E\ turns toward E 2 ), we must also 
require that 

co 12 (7) = V v Ei-E 2 . (*) 

These conditions completely determine V V W for any vector fields V and W: 

3.1 Lemma Assume that V is a covariant derivative on M with the usual 
linear and Leibnizian properties, and such that (*) holds for a frame 
field Ei,E 2 . Then V obeys the connection equations 

V v Ei = o) n (V)E 2 

V V E 2 = (aa(V)Ei. 

Furthermore if W = fiEi + fiE?. is an arbitrary vector field, then 

V r W = {V[fi] +/ 2 co 21 (F)}£ 1 + {V\M +/ 1 « a (F)}^,. 

We call this last expression the covariant derivative formula. Note that 
F[/i] and V\fi] only tell how W is changing relative to Ei,E 2 — the effect of 
the terms involving connection forms is to compensate for the way E x , E 2 
itself is rotating, so V Y W is an "absolute" rate of change. 

Proof. Since Ei'E 2 = 0, by a Leibnizian property of V, we get 

= V[.£a*.E(2] = VyE\»E 2 -f- E\ m ^yE 2 . 



Sec. 3] COVARIANT DERIVATIVE 319 

Hence, by (*), 

VyE 2 »E 1 = -wuOO = ton (7). 

But Ei*Ei = 1, so the same Leibnizian property gives 2V Y Ei*Ei = for 
i =■ 1, 2. Using this information, we derive the connection equations by 
orthonormal expansion of V v Ei and V V E 2 . 

Finally we apply the assumed properties of V to get 

= 7[/lR + /l V V ^! + 7[/ 2 ]#, + /2 V v # 2 . 

Substitution of the connection equations then gives the covariant deriva- 
tive formula. I 

This lemma shows how to define the covariant derivative of M. Note that 
the order of events is the reverse of that in Chapter II. There we used the 
Euclidean covariant derivative to define connection forms; now we shall 
use the connection form o>i 2 to define the covariant derivative for M. 

3.2 Theorem For a geometric surface M, there is one and only one co- 
variant derivative V with the usual linear and Leibnizian properties 
(Corollary 5.4 of Chapter II) and satisfying equation (*) for every frame 
field on M. 

Proof. The previous lemma shows that there is at most one such co- 
variant derivative, for V r W is given by a formula which does not involve 
V. So what we must prove is that such a covariant derivative V actually 
exists. The proof split into two parts, and we shall supress some details. 

A. Local definition. For a fixed frame field E lf E 2 on a region 0, use the 
formula in Lemma 3.1 as the definition of V r W. Routine computations 
verify that V is linear and Leibnizian, and when W is E lt we get 

V r Ei = un(V)E 2 ; 
hence (*) holds. 

B. Consistency. For two different frame fields, do the local definitions 
agree? If V V W derives from E u E% on 0, we must show that V r W = V V W 
holds on the overlap of and 0. For then we have a single covariant de- 
rivative on all of M. Because of the linear and Leibnizian properties, it 
suffices to show that 

V Y Ei = V F -Bi, VvE* = vVl?2. (1) 

We use Lemma 1.4, assuming for simplicity that E u E 2 and E u E 2 have 
the same orientation. Applying VV to the equation 



320 RIEMANNIAN GEOMETRY [Chap. VII 

E t = cos # Ei + sin # # 2 , 

the covariant derivative formula gives 

V v Ei = { F[cos #] + sint? co2i(F)}^i + { 7[sin#] + cost? o> n {V)}E 2 . (2) 

By Lemma 1.4, a>i 2 = coi 2 + d#. Substituting w 12 = a>i 2 — d& into (2) 
produces some favorable cancellations, leaving 

V Y E\ = a>a(V){ — sin & Ei + cos # # 2 } 

_ - - (3) 

= &i 2 (V)E 2 = VyE X . 

In the same way, V r E 2 = W-#2 may be derived from 

Ei = —sin # Ei + cos # Z? 2 . | 

3.3 Example The covariant derivative of E 2 . The natural frame field 
Ui, U 2 has coi2 = 0. Thus for a vector field 

W = hUx + ftU lf 

the covariant derivative formula (Lemma 3.1) reduces to 

V r W = V\Wi + V[f 2 ]U 2 . 

This is just Lemma 5.2 of Chapter II (applied to E 2 instead of E ), 
so our abstract definition of covariant derivative produces correct results 
on the Euclidean plane. 

The covariant derivative V of a geometric surface M may be modified 
so as to apply to a vector field Y on a curve a in M . (As usual, for each t, 
Y (t) is a tangent vector to M at a (t), as in Fig. 7.6. ) 

If Ei, Ei is a vector field on a region of M containing a, we may write 

Y(t) = y!(0^i(«(0) + y*(t)E,(a(t)), 
or, briefly, 

Y = yiEi + y 2 E 2 . 

Roughly speaking, we want the covariant derivative Y of Y to be V a ,Y. 
Thus the covariant derivative formula (Lemma 3.1) shows that we must 
define 

Y' = {yi + yxatiict )\Ei + {y 2 + yiun (a ))E 2 . 

It is a routine matter to check that this notion of covariant derivative is 
independent of the choice of frame field and has the same linear and 
Leibnizian properties as in the Euclidean case. Also, as in Chapter II, Sec- 
tion 2, since the velocity a of a curve in M is a vector field on M, we can 
take its covariant derivative to obtain the acceleration a." of a. 



Sec. 3] COVARIANT DERIVATIVE 321 




FIG. 7.5 

It may be well to look back now at the case of a surface M in E . If V 
and W are tangent vector fields on M, there are two ways to take covariant 
derivatives: one from the intrinsic geometry of M as a geometric surface, 
the other the Euclidean covariant derivative of E 3 . These two derivatives 
are generally different, but there is a simple relationship between them. 

3.4 Lemma Let V and W be tangent vector fields on a surface M in E 3 
(Fig. 7.5). If V is the covariant derivative of M as a geometric surface, 
and V is the Euclidean covariant derivative, then 

V r W is the component of y r W tangent to M. 

Proof. First suppose that W is one of the vector fields E lf E 2 of an adapted 
frame field E u E 2 E 3 . By the Euclidean connection equations (Theorem 
7.2 of Chapter II) we have 

vY#i = E <*a<y)Ei = <»n(V)E 2 + a> 13 (F)#3. 

3=1 

But the connection equations (Lemma 3.1) for M give 

VyEi = 0> 12 (V)E 2 . 

Thus VvEi is V v Ei plus a vector field normal to M. In other words, V v Ei 
is (at each point) the component of Vr-Ei tangent to M. The same result, 
of course, holds for E 2 . 

In the general case, since W is tangent to M, we may write 

W = fiE, + ftEt. 

Then the required result follows immediately from the special case above, 
since both covariant derivatives are linear and Leibnizian. | 

Thus we have been using the intrinsic covariant derivative of McE 
all along, although we did not give it formal recognition. It occurs when- 



322 RIEMANNIAN GEOMETRY [Chap. VII 




7(0) = V~ ^.p - a(0) 

FIG. 7.6 

ever we take the tangential component of the Euclidean covariant de- 
rivative. 

Only the most basic properties of covariant derivatives are shared by 
all geometric surfaces. In particular, the related notion of parallelism 
(due to Levi Civita) does not always behave as in the Euclidean case. 
Much of the individual character of Euclidean geometry rests on the fact 
that a tangent vector \ p to, say, E 2 may be moved to a parallel tangent 
vector v<j at any other point q. As we shall see, this phenomenon of "distant 
parallelism" does not obtain on an arbitrary geometric surface. However 
it is always possible to define parallelism of a vector field Y on a curve. In 
Euclidean space, this means that Y has constant coefficients relative to the 
natural frame field, but the infinitesimal characterization Y = makes 
sense in general. 

3.5 Definition A vector field Y on a curve a in geometric surface M is 
parallel provided its covariant derivative vanishes : Y' = 0. 

Just as in the Euclidean case, a parallel vector field has constant length, 
for || Y || 2 = Y-Y, and (Y*Y)' = 2Y-Y f = 0. 

3.6 Lemma Let a be a curve in a geometric surface M , and let v be a 
tangent vector at, say, p = a(0). Then there is a unique parallel vector 
field Fona such that 7(0) = v (Fig. 7.6). 

Proof. We may suppose that a lies entirely in the domain of a frame 
field Ei, Ei on M. (Otherwise we could break a up into segments for which 
this is the case.) The vector field 7 must satisfy the conditions 

V' = 0, 7(0) = v. (1) 

Because 7 has constant length || 7 || = c, we may write 

V = c cos <p Ei + c sin <p E 2 (2) 

where <p is the angle from E x to 7. Thus the covariant derivative formula 
gives 



Sec. 3] COVARIANT DERIVATIVE 323 

V' = c { — sin <p <p + sin <p co2i(a )} E x 
+ c {cos <p <p + cos <e> coi 2 (a )} #2 
It follows immediately that (1) is equivalent to 

<p = —oon(a ), 

with^>(0) the angle from E'i(p) to V(0) = v. There is only one such func- 
tion, namely 

<p(t) = (p(0) — I unia ) dt 
Jo 

This function <p, substituted in (2), defines the required vector field V. | 

In the situation stated in Lemma 3.6 we say, for each t, that the vector 
V(t) at a(t) is obtained from v at p = a(0) by parallel translation along a. 

In E 2 , parallel translation of a tangent vector y p along a curve segment 
from p to q merely produces the distant-parallelism result v q , which is 
thus entirely independent of the choice of the curve. But for an arbitrary 
geometric surface M, different curves from p to q will usually produce 
different vectors at q. Equivalently: // a vector vatpis parallel-translated 
around a closed curve a (starting and ending at p) the result v is not neces- 
sarily the same as v. This phenomenon is called holonomy. If we fix a par- 
ticular frame field on the curve a, then the proof of Lemma 3.6 shows that 
parallel translation from a (a) = p to a(b) = p along a rotates all vectors 
through the same angle <p(b) — <p(a) — since <p is the same for all parallel 
vector fields. We call this the holonomy angle yp a of a. (Multiples of 2r 
may be ignored in \f/ a , since they do not affect the determination of v*.) 

3.7 Example Holonomy on the sphere 2 of radius r. Suppose the closed 
curve a parametrizes a circle on 2. There is no loss of generality in assum- 
ing that a is a circle of latitude, say the w-parameter curve 

a(u) = x(u,v ), ^ u ^ 2x, 

where x is the geographical patch in 2 (Fig. 7.7). According to Example 
6.2 of Chapter VI, the associated frame field E u J£ 2 of x has o>n = sin v du. 
Now the proof of Lemma 3.6 shows that every parallel vector field on a 
has angle p (measured from #1) satisfying <p — — com (a ). It follows that 
<p has constant value —sin vq on a. Thus the holonomy angle \f/ a of a is 

<p(2ir) — <p(0) = — 2-k sin v 

Note that only on the equator, v = 0, does a vector v return to itself 
after parallel translation around a. When Vo is near ir/2, then a is a small 
circle around the north pole of 2. Since <p is close to —1, parallel vector 
field V is rotating rapidly with respect to Ei, E 2 . But the holonomy angle is 



324 



RIEMANNIAN GEOMETRY 



[Chap. VII 



T,(2) 




FIG. 7.7 

close to — 2t, so the actual difference between v = V(0) and v* = V(2ir) 
is, as one would expect, quite small. 

Gaussian curvature has a strong influence on holonomy, as shown by 
Exercise 5. 

For a patch x in an arbitrary geometric surface M we shall inevitably 
use the notation x uu for the covariant derivative of x u along the w-parame- 
ter curves of x — with corresponding meanings for x uv , x„, and x vv . Thus 



when M is a surface in E , we must have a new notation, say x. 



, for 



the analogous objects defined in Chapter V, Section 4. There we were 
using the covariant derivative of E 3 , while now we use the covariant de- 
rivative of M itself. It is still true in the intrinsic case that x uv = x vu , 
but the proof is by no means obvious (Exercise 9 ) . 



EXERCISES 

1. In the Poincare' half -plane, let a be the curve given in Exercise 2 of 
Section 1. Express its velocity and acceleration in terms of the frame 
field 

Ei = vUi, E 2 = vU 2 . 

(Hint: They are collinear.) 

2. Let j8 be the curve 

|8(0 = (ct,st),t > 0, 

in the Poincare" half -plane, where c and s are constants with c 2 + s 2 = 1. 
(Thus /3 is a Euclidean straight line through the origin.) Express the 
velocity and acceleration of in terms of the frame field used in Exer- 
cise 1. 

3. If V and W are tangent vector fields on a surface in E 3 , refine the proof 
of Lemma 3.4 to show that 



Sec. 3] COVARIANT DERIVATIVE 325 

V V W = VyW + S(V)-WU, 

where S is the shape operator derived from U = ztE 3 . Hence if a is a 
curve in M, 

a" = a." + S(a)-a U. 

4. Show that on the sphere 2 the curve a given in Example 3.7 has (in- 
trinsic) acceleration a" = r cos Vo sin vo E 2 - Compute its Euclidean 
acceleration, and show that a" is the component tangent to 2. 

5. Let a be a closed curve in a geometric surface M. 

(a) If a is homotopic to a constant via x (Ex. 12 of IV.6), show that 
the holonomy angle of a is J/ x K dM. (Assume that x(R) lies 
in the domain of a frame field.) When x is patch-like, this inte- 
gral is the total curvature of the region x(R.) 

(b) Compute the holonomy angle in Example 3.7 by this method. 

6. Let V be a parallel vector field on a curve a in M, and let W be a vector 
field on a with constant length. Show that W is parallel if and only if 
the angle between V and W is constant. 

7. Show that isometries preserve covariant derivatives in this sense: If 

Y is a vector field on a curve a in M, and F: M — * M is an isometry, 
then 

F* (Y f ) = ?', where Y is the vector field F* (F) on a = F (a) in Af. 

(Simplify matters by assuming that Y can be written Y = fE u where 
Ei, Ei is a frame field on M). 

This is the analogue of the Euclidean result (Corollary 4.1 of Chapter 
III). The general case is given in Exercise 8. 

8. Prove that an isometry F.M-+M preserves the covariant derivatives 

V and V. Explicitly, for each vector field X on M , let X be the trans- 
f ered vector field o n M: X(F(p)) = F*(X (p )) for each point of M .Then 
show that VyW = V#. (Hint: If /< = W *Ei and /» = W-E h use Exer- 
cise 8 of Section 5, Chapter IV, to show that V[fi] at p equals V[fi] at 
F(p).) 

9. If x is an orthogonal patch in a geometric surface M, show that 

(intrinsic derivatives). (Hint: For the associated frame field; compute 
/ \ ( Xv \ Zb* x " 

«*(*.) = Vv^A V^ " Xvu Weg 

Then using the formula for w n given in Chapter VI, Section 6, show that 



326 RIEMANNIAN GEOMETRY [Chap. VII 

W21 ( xtt )=__ = xttv __. 



Find an easier proof in the special case where M is a surface in E 3 . 

10. (Continuation) Show that y uv = y vu for an arbitrary patch y. (Hint- 
There exists an orthogonal patch x such that x = y(u,v).) 

11. If there exists a nonvanishing vector field W on M such that V V W = 
for all V, show that M is flat. Find such a vector field on a cylinder 
in E 3 . 



4 Geodesies 

Geodesies in an arbitrary geometric surface generalize straight lines in 
Euclidean geometry. We have seen that a straight line a(t) = p + tq is 
characterized infinitesimally by vanishing of acceleration; thus 

4.1 Definition A curve a in a geometric surface M is a geodesic of M 
provided its acceleration is zero; / = 0. 

In other words, the velocity a of a geodesic is parallel: geodesies never 
turn. Recall that a parallel implies \\a || constant; so geodesies have 
constant speed. 

Because acceleration is preserved by isometries (Exercise 7 of Section 
3), it follows that geodesies are isometric invariants. (A direct proof 
appears in Exercise 1 of Section 5, Chapter VI). In fact, if F: M — > N is 
merely a local isometry, then F carries each geodesic a of M to a geodesic 
F(a) of N — for F is locally an isometry, as discussed in Chapter VI, Sec- 
tion 4. 

The general definition of geodesies given above agrees with that of 
5.7 in Chapter V when M is a surface in E 3 , for we can deduce from Lemma 
3.4 that the intrinsic acceleration of a curve ainMcE is the component 
tangent to M of its Euclidean acceleration. Thus the former is zero if 
and only if the latter is normal to M. 

Suppose that a: I — » M is a curve in an arbitrary geometric surface M 
and Ei, E 2 is a frame field on M . Throughout this section we use the nota- 
tion 

a = ViEi + v 2 E<t and a" = A^i + A 2E2 

for the velocity and acceleration of a. From Section 3 we know that these 
components of acceleration are 

A x = Vi + V 2 a)2i(ot') 
A 2 = v 2 ' + Viwn(a) 



Sec. 4] GEODESICS 327 

and are real-valued function on the interval I. Our main criterion for a 
to be geodesic is thus A x = A 2 = 0. Using orthogonal coordinates we now 
rewrite these equations in a more informative way. 

4.2 Theorem Let x be an orthogonal coordinate patch in a geometric 
surface M. A curve a{t) = xfait), a*(t)) is & geodesic of M if and only if 

ai " + hi {Euai ' 2 + 2E ^ a * " aa2 ' 2} = ° 

a 2 " + ~ \-E v a x 2 + 2(? M a 1 V -f G v a 2 2 } = 0. 

We shall subsequently refer to these equations as A x = and A 2 = 0. 
Note that they are symmetric, in the sense that the reversals 1 <-> 2, u <-> v, 
E <-> G turn each one into the other. In this context we shall always under- 
stand that the functions E, G and their partial derivatives E u , E v , • • • are 
evaluated on (a x , o 2 ), and hence become functions on the domain J of a. 

Proof. The velocity of a is a = o/x u + a 2 x v , so in terms of the asso- 
ciated frame field of x (Chapter VI, Section 6), we have 

«' = (a x VE)E X + (a 2 VG)E 2 . 

Thus the acceleration components A x , A 2 defined above become 

A x = (a x VE)' + (a 2 ' VG)<*i(a) 

A 2 = (a 2 ' \/G)' + (a x ' VE)o>n(d). 
Using the formula for a>i 2 from Chapter VI, Section 6, we find 

/ '\ t ' i ' ^ ('vE)v „ i , ('vG)u i / 9 \ 
o>n{<* ) = wi 2 (aix„ + a 2 xj = ^=- ai + — -7=- a 2 U; 

When this is substituted in (1), the geodesic equations A x = A 2 = 
become 

(a/ VEY + (VEWa.' " V ^ ( ^ )m a/ 2 = 



(1) 



(a,' VG)' - ^V^ «i' 2 + (VG)4h'<h = 0. 



(3) 



Standard calculus computations will transform (3) to the form stated in 
the theorem. We merely remind the reader that in a Leibnizian expansion 
such as 

77" 

(a x VE)' = «i" VE + a x J^= , 



328 RIEMANNIAN GEOMETRY [Chap. VII 

the notation E is short for E(a 1} (h), so 

E' = E u ai + E v a 2 . | 

4.3 Theorem Given a tangent vector v to I at a point p, there is a 
unique geodesic a of M such that 

a(0) = p, a'(0) = v. 

Thus there are lots of geodesies in any geometric surface, and each is 
completely determined by its initial position and velocity. In E 2 for ex- 
ample, the geodesic determined by v at p is the straight line a(t) = p + t\. 

Proof. Let x be an orthogonal patch in M with p = x(u , v ), and write 
v = cox u + d x v . The geodesic equations in Theorem 4.2 have the form 

a/' = /i (ai, a 2 , ai , a 2 ) 

(h = j2 (ai, «2, ai , «2 ) 

Furthermore, a will satisfy the given initial conditions if and only if 

ai(0) = wo a/(0) = Co 

a 2 (0) = vo a 2 ' (0) = d . 

Now the fundamental existence and uniqueness theorem of differential 
equations asserts that there is an interval / about on which are defined 
unique functions a u a 2 which satisfy (1) and (2). Thus a = x(ai, a 2 ) is 
the only geodesic defined on / such that a(0) = p, c/(0) = v. | 

This proof is not entirely satisfactory, because the interval / may be 
unnecessarily small. We describe briefly a way to make it as large as possi- 
ble. Suppose that ct\: Ii — > M and a 2 : h —> M are geodesies satisfying the 
same initial conditions at t = 0. Using the uniqueness property above, we 
can deduce that ai = a 2 on the common part of I x and 7 2 . Applying this 
consistency result to all such geodesies, we obtain a single maximal geodesic 
a: I — * M satisfying the initial conditions. (The interval / is the largest 
possible.) Intuitively this means we simply let the geodesic run as far as 
it can. 

4.4 Definition A geometric surface M is geodesically complete provided 
every maximal geodesic is defined on the whole real line R. 

Briefly: geodesies run forever. A constant curve is trivially a geodesic, 
but excluding this case, every geodesic has constant nonzero speed. Thus 
geodesic completeness means that all nontrivial (maximal) geodesies are 
infinitely long — in both directions. For example, E 2 is certainly complete, 
and the explicit computations in Example 5.8 of Chapter V show that 



Sec. 4] GEODESICS 329 



N = J(T) 




FIG. 7.8 

spheres and cylinders in E 3 are. More generally, all compact geometric 
surfaces are complete, as are all surfaces in E 3 of the form M: g = c (con- 
sequences of Theorem 15, Chapter 10, of Hicks [5]). Removal of even a 
single point from a complete surface will destroy the property since geo- 
desies formerly passing through the point will be obliged to stop. 

There is a Frenet theory of curves in a geometric surface M which 
generalizes that for curves in the plane (Exercise 8 of Section 3 in Chapter 
II). Because M has only two dimensions, torsion cannot *oe defined. If M 
is oriented, however, curvature can be given a geometrically meaningful 
sign, as follows. If 0: / — > M is a unit-speed curve in an oriented geometric 
surface, then T = fi' is the unit tangent vector field of 0. To get the principal 
normal vector field N, we "rotate T through +90°," denning 

N = J(T), 

where J is the rotation operator from Exercise 5, Section 1 (Fig. 7.8). 
Then the geodesic curvature k„ of is the real-valued function on / for which 
the Frenet formula T' = k N holds. Thus k is not restricted to nonnegative 
values as in the case of curves in E 3 : k > means that T — hence 0— is 
turning in the positive direction as given by the orientation of M, and 
Kg < means negative turning. 

4.5 Lemma Let be a unit-speed curve in a region oriented by a frame 
field E u Ei. If <p is an angle function from Ei to #' along /3, then 

Kg = j — h 0>l 2 (j3 ). 

ds 

Proof. By definition of angle function (Exercise 15 in Chapter VI, Sec- 
tion 7), we have 

T = 0' = cos <p Ei + sin <p E 2 . 

Since the orientation derives from this frame field, 

J(E X ) = E 2 and J (E 2 ) = -E lt 

so 



330 RIEMANNIAN GEOMETRY [Chap. VII 

N = J(T) = -sin <pEi + cos v E 2 . 
Using the derivative formula on page 320, we find 

T = /3" = { — sin tp <p' + sin <p wsiOs')}^ 
+ {cos <p <p f + cos ^ wbOs')}^. 
But K g = T'-N = T'-J(T), so using the formulas for T' and /(T 7 ), we get 
K g = (cosV + sinV)0' + coi 2 (/3')) = */ + wiaOS 7 ). | 

For example, in E 2 the natural frame field has o>i 2 = 0, and <p becomes 
the usual slope angle of the curve 0. Thus the result reduces to k = a\p/ds, 
which in elementary calculus is often taken as the definition of curvature. 

For an arbitrary-speed regular curve a in M, the Frenet apparatus 
T, N, K g is defined — just as in Chapter II, Section 4 — by reparametrization. 
Furthermore the same proof as for Lemma 4.2 of Chapter II shows 



dv 
di 



a = vT a." =%T + Kg v 2 N (*) 



where v = \\ a \\ is the speed function of a. 

4.6 Lemma A regular curve a in M is a geodesic if and only if a has 
constant speed and geodesic curvature k = 0. 

Proof. Since v > 0, we have a" = if and only if (dv/dt) = k = 0. | 

/ 

The equations ( ) also show that a has geodesic curvature zero if and only 
if a and a" are always collinear. Such curves are sometimes called geo- 
desies: to get a geodesic in the strict sense of Definition 4.1 it suffices to 
reparametrize a to a constant speed curve. (Proof: k is unaffected by 
reparametrization.) In contexts where parametrization is of some im- 
portance we shall call a curve with K s = 0a pre-geodesic. 

Computation of explicit formulas for the geodesies of a given geometric 
surface is rarely a simple task. Our purpose, however, is not to collect 
formulas, but to study the general behavior of geodesies. Before continuing 
this study we are going to examine an important special case in which a 
considerable amount of concrete information about geodesies can often 
be obtained with only a minimum of computation. 

4.7 Definition A Clair ut parametrization x: D —*■ M is an orthogonal 
parametrization for which E and Cr depend only on u, that is, F = and 
E v = G v = 0. 

For example, the usual parametrization of a surface of revolution is a 
Clairut parametrization. 

4.8 Lemma If x is a Clairut parametrization, then 



Sec. 4] GEODESICS 331 

1. All the n-parameter curves of x are pre-geodesics, and 

2. A ^-parameter curve, u = u , is a geodesic if and only if G„(w ) = 0. 

Proof. For (1) it suffices by a remark above to show that x u and x uu 
are collinear. Since x„ and x„ are orthogonal, this is equivalent to 

X V 'X UU = U. 

But the following equations imply this result: 

= tLv = (.Xii^XtiJu == ZX U *X UV 

= fr u = \x u *x v ) u == x uu *x v ~r x u *x vu . 

Similarly, for (2), the v-parameter curve, u = u , is pre-geodesic if and 
only if x vv (u , v)'x u (u , v) = 0. The following equations show that this is 
the case if and only if G u (uq) = 0. 

U = f j = Xtiv * X v ~j~ X u • X vv 

G u (uo) = G u (u , v) = 2x vu (u , v)»x v (uo, v) for all v. 

(Recall that x uv = x vu .) We have not used the condition G v = 0; its only 
effect is to show that y-parameter pre-geodesics are in fact geodesies, since 
it means that v-parameter curves have constant speed. I 

In the case of a surface of revolution this lemma provides an intrinsic 
proof that meridians are geodesies and that a parallel, u = u , is geodesic 
if and only if ti (u ) — 0. (See Exercise 3 of Chapter V, Section 5). 

Because of the preceding lemma we shall think of a Clairut parametriza- 
tion as a "flow" whose streamlines are its w-parameter geodesies, and we 
shall measure the behavior of arbitrary geodesies relative to this flow. 

4.9 Lemma If a = x(a u a^) is a unit speed geodesic, and x is a Clairut 
parametrization, then the function 

c = G(ai)ch = s/G(ai) sin <p 

is constant, where ip is the angle from x„ to a . Hence a cannot leave the 
region for which G ^ c . 

We call the constant c thus associated with each geodesic a the slant 
of a, since^ — in combination with G — it determines the angle <p at which a 
is cutting across the w-parameter streamlines of x (see Fig. 7.9). 

Proof. Because E v = G v = for a Clairut parametrization, the equation 
A 2 = of Theorem 4.2 reduces to 

a 2 + y7 ai a 2 =0 



332 



RIEMANNIAN GEOMETRY 



[Chap. VII 




FIG. 7.9 



But this is equivalent to the constancy of c = G(h, since 
(Gat')' = G' a* + G a 2 " = GUi' a,' + G a 2 " 
To show that c = -y/G sin <p, compare the two equations 

a 'X v = (ai x u + a 2 x„)»x r = (ja 2 ' = c 

a'-x, = || a || || x„ || cos (f "~ ^) = V& sin <p 
It follows immediately from | sin <p | ^ 1 that 6r ^ c 2 . 



I 



If a is moving in a direction in which G is increasing, then the constancy 
of c = y/G sin <p shows that ^> is decreasing: a is forced to turn more in the 
direction of the flow. On the other hand, if G is decreasing along a, then 
a cuts across the w-parameter geodesies at ever-increasing angles. Some 
interesting consequences are presented in Exercises 11 and 12. This inter- 
pretation is particularly simple on a surface of revolution (Exercise 13). 

We can add to the lemma above the equation 



ai = =fc 



\/G - i 

Veg 



(1) 



In fact, since a has unit speed we have 1 = «'•«' = Zfa/ 2 + Ga 2 ' 2 . In this 
equation substitute 



a 2 = 



G 



(2) 



(from Lemma 4.9), and solve for a/ to obtain equation (1). 

Conversely, a straightforward computation shows that if a/ is nonzero, 
the equations (1) and (2) imply that a = x(ai, a*) is a unit speed geo- 



Sec. 4] GEODESICS 333 

desic. Furthermore, a/ nonzero is a necessary and sufficient condition for an 
arbitrary curve a to have a reparametrization of the form 

j8(w) = x(w, v(u)). 

The point of all this is that we can now give a relatively simple criterion 
for a curve of this form to be a pre-geodesic — and thereby determine the 
routes of the geodesies in a region with a Clairut parametrization. (The 
special parametrization of /3 misses essentially only the geodesies given by 
(2) of Lemma 4.8; see Exercise 12.) 

4.10 Theorem A curve /3(w) = x(u, v(u)), where x is a Clairut para- 
metrization, is a pre-geodesic if and only if 

— = ±c\/E 

d^ ~ VGVG - c 2 
The constant c is then the slant of /3. 

Proof. The argument is simply an exercise in change of parametrization. 
Let a be a unit speed reparametrization of p derived just as in Chapter II 
from an arc length function s for 0. Thus is pre-geodesic if and only if 
a is geodesic. Let a,\ be the inverse function of s (so Oi does not vanish). 
Then 

a = (8(ai) = x(ai, y(«i)), 

and we set a% = v{a\). The remarks above show that a is a geodesic (of 
slant c) if and only if 

a\ = — _T c , a 2 ' = -^ (E,G evaluated onci) (1) 
y/EG " 



If these equations hold, then by elementary calculus 

dv _ azis) _ ±c \/E 
du ~ ax'(s) ~ y/Q VG - c 2 



(2) 



where the substitution of s (inverse function of ai) makes E and G merely 
functions of u. Conversely, if (2) holds, we deduce (1) using the equation 
Ea x ' 2 + Go* 2 = 1 which expresses the fact that a has unit speed. | 

Since the formula above for dv/du depends only on u, by the fundamental 
theorem of calculus it can be written in integral form as 



v(u) = v(uo) ± / , nW7T - 



'o Vg Vg - c 2 

Thus for a Clairut parametrization — in particular, for a surface of revo- 



334 RIEMANNIAN GEOMETRY [Chap. VII 

lution — the computation of pre-geodesics is reduced to a single integration. 
This is, of course, a vastly simpler criterion than the second-order differ- 
ential equations i)f Theorem 4.2. Unfortunately, however, the integration 
can rarely be carried out in terms of elementary functions. 

4.11 Example Routes of Geodesies 

(1) The Euclidean plane E . We begin with a surface whose geodesies we 
already know, but to illustrate the preceding result we shall find their 
routes in terms of the polar parametrization 

x(u, v) = (u cos v, u sin v). 

Since E — 1, F = 0, and G = w 2 , this is a Clairut parametrization. The 
w-parameter geodesies are just the radial lines through the origin. All 
others may be parametrized as /5(w) = x(u, v(u)), where by Theorem 4.10 

dv_ = ±c ^ d / _ x c\ 

du u \Zu 2 — c 2 du\ u) 

Hence v — v = ± cos -1 (c/w), or u cos (v — v ) = c, which is the polar 
equation of a straight line. The slant c has geometrical significance as the 
distance from the line to the origin. 

(2) The hyperbolic plane H. Polar coordinates are a more natural choice 
in this case since the function g giving the geometric structure of H de- 
pends only on distance to the origin, f Thus if 

x(u, v) = (u cos v, u sin v), < u < 2, 

thengf(x) = 1 — w 2 /4. (We write simply g henceforth.) Now 



1 u 

E = x„ ° x„ = — , F = 0, G = x„ o x „ = — ; 

r g 2 

thus x is a Clairut parametrization. By Lemma 4.8 the w-parameter curves 
— Euclidean lines through the origin — are routes of geodesies of H. By 
Theorem 4.10, /3(w) = x(w, v (u)) is pre-geodesic provided 

<fo _ ±(cg/u 2 ) 

du Vl - (cg/u) 2 ' { ) 

To carry out the required integration, set 

w = -\1 -\- — }, where a = 
u\ 4/ 



Vl + c 2 " 
Then a straightforward computation yields 

t An even better choice, the hyperbolic polar coordinates discussed in Example 
5.5, gives rise to the substitution used later in this example. 



Sec. 4 



GEODESICS 



335 



dv 
du 



T dw/du 
Vl - w 2 



(2) 



Hence 



Thus 



v — v = ±cos w, or cos (v — Vo) = w = - 



4w 



-SO + s)- 



u + 4 cos (v — Vo) = 

a 



(3) 



Using the law of cosines in a diagram similar to that in Fig. 7.10, one finds 
the polar equation of a circle of radius r, centered at x(u , v ). 



2 , 2 

U + Uq 



2u u cos (y — v ) = r 2 . 



(4) 



Comparison with equation (3 ) shows that the route C of /3 is a Euclidean 
circle with wo 2 — r 2 = 4. Since w > 2, the center of C lies outside the 
hyperbolic plane H: x 2 + y < 4. One can see from Fig. 7.10 that the circle 
C is orthogonal to the rim x 2 + y 2 = 4 of #. Of course, lies in the open 
arc of C inside H, and we deduce from Theorem 4.2 that fills this arc. 

Conc/us/on. The routes of the geodesies of the hyperbolic plane H are 
the portions in H of : all Euclidean straight lines through the origin, and all 
Euclidean circles orthogonal to the rim of H. 

The argument in Example 2.4 suggests that the geodesies of H have 
infinite length (a formal proof is given in Exercise 1 of Section 5); thus 
H is geodesically complete. 

The geodesies of the hyperbolic plane bear comparison with those of 
the Euclidean plane. Around 300 B.C. Euclid established a celebrated 
set of axioms for the straight lines of his plane. The goal was to derive its 




x(w , t'o) 



FIG. 7.10 



336 RIEMANNIAN GEOMETRY [Chap. VII 




FIG. 7.11 

geometry from axioms so overwhelmingly reasonable as to be "self-evi- 
dent." The most famous of these is equivalent to the 'parallel postulate: 
If p is a point not on a line a, then there is a unique line through p which 
does not meet a. Over the centuries this postulate began to seem some- 
what less self-evident than the others. For example, the axiom that two 
points determine a unique straight line might be checked by laying down 
a (perhaps long, but still finite) straight edge touching both points. But 
for the parallel postulate one would have to travel the whole infinite length 
of j8 to be sure it never touches a. Thus tremendous efforts were expended 
in trying to deduce the parallel postulate from the other axioms. The 
hyperbolic plane H offers the most convincing proof that this cannot be 
done. For if we replace "line" by "route of a geodesic," then every Euclid- 
ean axiom holds in H except the parallel postulate. For example, given 
any two points it is easy to see that one and only one geodesic route runs 
through them. But it is clear from Fig. 7.11 that in H there are always an 
infinite number of geodesic routes through p that do not meet a. When the 
implications of this discovery were worked out, what was destroyed was 
not merely the modest hope of deducing the parallel postulate, but the 
whole idea that E 2 is, in some philosophical sense, an Absolute, whose 
properties are "self-evident." It had become but one geometric surface 
among the infinitely many others discovered by Riemann. 



EXERCISES 

1. Show that a reparametrization a(h) of a nonconstant geodesic a is 
again a geodesic if and only if h has the form h(t) = at + b. 



Sec. 4] GEODESICS 337 




FIG. 7.12 

2. Denote by y v the unique geodesic in M with initial velocity v. For 
any number a, show that y av (t) = y v (at) for all t. 

3. Let V be a vector field on a geodesic a. Show that V is parallel if and 
only if || V || is constant and V makes a constant angle with a . 

4. In the sphere 2, let nbe the north pole, pi and p2 points on the equator. 
Consider the broken curve /3 following a meridian from n to p x , the 
equator from pi to p2, then a meridian from p2 back to n. Prove that 
the holonomy angle of /3 is the angle at n between the two meridians. 

5. Find the routes of the geodesies in the stereographic sphere, (1) of 
Example 2.3. 

6. In the Poincare" half -plane, show that the routes of the geodesies are: 
all semicircles with centers on the w-axis, and all vertical lines. (Hint: 
x (w, v) = (u, v) is a Clairut patch "relative to v," so in the text 
equations, reverse u and v, and E and G.) See Fig. 7.12. 

7. Let a be a unit speed curve such that a is never collinear with Ei, 
where E u Ei is a frame field. If a" = AiEi + A 2 E 2 , show that the 
single equation A\ = implies that a is a geodesic. 

8. In the projective plane of radius r (Exercise 6 of Section 2), prove: 

(a) The geodesies are simple closed curves of length irr. 

(b) There is a unique geodesic route through any two distinct points. 

(c) Two distinct geodesic routes meet in exactly one point. 
(Hint: every geodesic in 2 is the image under the projection P: 
2 — > S of a geodesic in the sphere 2.) 

9. If a is a curve in M with speed v > 0, prove: 

(a) The geodesic curvature n g of a is a" *J (a )/v . Hence if M is a 
surface in E 3 , k = U-a X a." /v. 

(b) If a has unit speed, then in Ex. 7 of Chapter V, Section 5, the 
vector field V is the unit normal N of a, and the function g is 
geodesic curvature k . 

10. Let M be the plane with the origin deleted, and furnish M with the 

conformal geometric structure for which g = r = \/u 2 -f- v 2 . 
Find the Gaussian curvature of M and the routes of its geo- 
desies. Show that M is isometric to a surface in E . 

11. Let x be a Clairut parametrization, and let a = x(ai, a^) be a unit 



338 



RIEMANNIAN GEOMETRY 
U = tt, 



[Chap. VII 




w-parameter curves 



FIG. 7.13 

speed geodesic with slant c. Suppose that a starts at the point 

a(0) = x(«i(0), a 2 (0)) = x(u ,v ) 

and that ai (0) > 0. If there is a number u > u such that G(u) = 
c, let Wi be the smallest such number. Then the v-parameter curve 

0(v) = x(wi, v) 

is called a barrier curve for a. Prove that: 

(a) a comes arbitrarily close to /3 

(b) If j8 is a geodesic, a does not meet /S (thus a asymptotically ap- 
proaches/8). 

12. (Continuation) If the barrier curve /3 is not a geodesic, it can be shown 
thatadoesmeetjS.flf a (£*) is the meeting point, show that cti (t*) = 
and that a± changes sign at t*. Thus a bounces off fi as shown in Fig. 
7.13. (Hint: prove that af (t*) < 0.) 

13. Let a be a geodesic on a surface of revolution. 

(a) Show that the slant of a is c = h sin <p, where h(t) is the distance 
from a(t) to the axis of revolution, and <p gives the angles at 
which a cuts the meridians of M . 

(b) Deduce that a cannot cross a parallel of radius \c\. 

1 4. Let a be a geodesic of slant c on the paraboloid of revolution 

M: z = x + y 2 - 

Find the minimum value of z(a), that is, the lowest height to which 
a descends. (Hint: Use a Monge patch.) 

15. Prove that no geodesic on the bugle surface (V.6.6) can be defined 
on the whole real line. 

16. On a torus of revolution, let a be a geodesic that at some point is 
t We assume, of course, that a remains in the region parametrized by x. 



Sec. 5] LENGTH-MINIMIZING PROPERTIES OF GEODESICS 339 

tangent to the top circle (u = ir/2). Show that a remains always on 
the outer half of the torus (— t/2 ^ u S t/2) and travels around 
the torus oscillating between the top circle and bottom circle. 

17. Let C be a catenoid (Example 6.1 of Chapter V) with c = 1, and let 
a be the geodesic such that 

a(0) = x(wo, Vo), Wo 5^ 0, 

and a (0) makes angle <p with the meridians. (Note that <p and ir — <po 
determine different parametrizations of the same geodesic.) For which 
values of <p does a cross the minimal circle, u = 0, of C? 

18. A Liouville parametrization x: D — » ilf is an orthogonal parametriza- 
tion for which E = G = U + V, where £7 is a function of u only and 
7 a function of v only. If a — x(a u ch) is a unit speed geodesic ex- 
pressed in terms of such a parametrization, show that 

Uish) sin 2 (p — V((h) cos 2 <p 

is a constant, where <p is the angle from x u to a . 

19. Let Ei, E 2 be a frame field on a geometric surface M. For i = 1, 2> 
let *»(p) be the geodesic curvature at p of the integral curve of E< 
through p. 

(a) Prove that K = Ei[k 2 ] - E 2 [ki] - k x 2 - k 2 \ 

(b) Test this formula on an arbitrary surface of revolution, using 
the frame field in Example 6.4 of Chapter VI. 

(Hint: for (a), prove coi 2 (i?i) = k») 



5 Length-Minimizing Properties of Geodesies 

The previous section considered geodesies as straightest curves; now we 
investigate their character as shortest curves. The basic problem is, roughly 
speaking, to find the shortest route from one point to another in a geo- 
metric surface. For E 2 the solution is simple: Given any two points p and 
q, there is a unique straight-line segment from p to q, and this is shorter 
than any other curve from p to q (Exercise 11 of Section 2 of Chapter II). 
For an arbitrary geometric surface M, the situation is more interesting. 
In the first place, there may be no shortest curve from p to q (Exercise 3 
of Section 4 of Chapter VI). And even if there is one, it may not be unique. 
For example, we shall soon prove the expected result that on a sphere, all 
semicircles from the north pole to the south pole have the same shortest 
length. To make the terminology precise, we use the notion of intrinsic 
distance (Chapter VI, Section 4). 



340 



RIEMANNIAN GEOMETRY 



[Chap. VII 



5.1 Definition Let a be a curve segment from p to q in M . Then 

(1) a is a shortest curve segment from p to q provided L(a) = p(p, q) 

(2) a is the shortest curve segment from p to q provided that 

L(a) = p(p, q) 

and that any other shortest segment from p to q is merely a reparametriza- 
tion of a. 

In the first case we shall also say that a minimizes arc length from p to q; 
the definition means that if fi is any other curve segment from p to q, then 
L(fi) * L(a). 

In the second case, we say that a uniquely minimizes arc length. "Unique- 
ness" must be interpreted liberally enough to allow for reparametrization, 
since monotone reparametrization (Exercise 10 of Section 2 of Chapter II ) 
does not change arc length. 

All such shortest curves will turn out to be geodesies (Lemma 5.8). 
Our first main result (Theorem 5.6) will show that short enough geodesic 
segments behave as well in an arbitrary geometric surface as they do in E 2 . 
Some preparatory work is needed first. 

In the Euclidean plane, if one is interested in the distance to the origin, 
it is natural to use polar coordinates, for then the distance from to 

x(u, v) = (u cos v, u sin v) 

is simply u. We shall now generalize this parametrization to the case of an 
arbitrary geometric surface M . As for E 2 , the w-parameter curves will be 
geodesies radiating out from some fixed point p of M. Such geodesies may 
conveniently be described as follows: If v is a unit tangent vector at p, 
let 7 V be the unique geodesic which starts at p with initial velocity v. Now 
we assemble all these geodesies into a single mapping as follows: 

5.2 Definition Let e x , e 2 be a frame at the point p of M . Then 



x(u, v) = 7, 



cos tie, + sin 



•e,(w) 




FIG. 7.14 



is the geodesic polar mapping of M 
with pole p. 

Here the domain of x is the larg- 
est region of E on which the for- 
mula makes sense. A choice of v fixes 
a unit tangent vector 

v = cos vei + sin vez 

at p (Fig. 7.14). Then the w-param- 



Sec. 5] 



LENGTH-MINIMIZING PROPERTIES OF GEODESICS 



341 



eter curve 



x(w, v) = y r (u) 



is the radial geodesic with initial velocity v. Since || v || = 1, this geodesic 
has unit speed, so that the length of y v from p = 7 V (0) to y y (u) is just u. 
In the special case where ei, *% is the natural frame at the origin of E 2 , 
the geodesic polar mapping becomes 

x (u, v) = 7 cos „ ei + sin VBi (u) 

= + w(cos vei + sin Vd) 

= (u cos v, u sin v). 

Thus x is a generalization of polar coordinates in the plane. 

The pole p is a trouble spot for a geodesic polar mapping. To clarify the 
situation near p, we define (in the situation described in Definition 5.2) 
a new mapping 

y(u,v) = Tuei +»e 2 (l)- 

Differential equations theory shows that y is differentiable, and it is easy 
to check that y is regular at the origin. Thus by the inverse function the- 
orem, y is a diffeomorphism of some disc D t : u 2 + v < i onto a neighbor- 
hood dl f of p. We call 9l £ a normal neighborhood of p. In the special case 
M = E 2 , y is just the identity map y(u,v) = (u,v),so for arbitrary M, y 
is a generalization of the natural (rectangular) coordinates of E 2 . 

5.3 Lemma For a sufficiently small number e > 0, let S t be the strip 
< u < e in E 2 . Then a geodesic polar mapping x: S t — > M with pole p 
parametrizes a normal neighborhood 3l t of p — omitting p itself, (see Fig. 
7.15). 



—s t - 




FIG. 7.15 



342 RIEMANNIAN GEOMETRY [Chap. VII 




FIG. 7.16 

Proof. Note that x bears to y the usual relationship of polar to rectan- 
gular coordinates; that is, 

X (U, V ) = y C os vei + sin VB2 V^/ = Tu cos i-ei + m sin »e2 \ / 

= y(w cos v, u sin v) 

where we use the identity y v (u) = 7« v (l) from Exercise 2 of Section 4. 
Now this formula expresses x as the composition of two regular mappings: 

(1) The Euclidean polar mapping (u, v) — > (u cos v, u sin v) which 
wraps the strip S t around the disc D t , and 

(2) The one-to-one mapping y of D t onto 9l e . 

Thus x is regular and carries S t in usual polar-coordinate fashion onto the 
neighborhood 9l £ — omitting only the pole. | 

We draw a fundamental consequence: If q = x(wo, Vo) is any point in a 
normal neighborhood 91 of p, then there is only one unit speed geodesic from 
p to q which lies entirely in 91, namely, the radial geodesic 

y(u) = x(u, v ) ^ u ^ uq. 

(Proof: Any unit-speed geodesic starting at p is, by the uniqueness of 
geodesies, a w-parameter curve of the polar parametrization. As suggested 
by Fig. 7.16, all except v = v + 2im lead out of 91 without hitting q, and 
different choices of n still give the same geodesic y by the usual ambiguity 
of polar coordinates). 

5.4 Lemma For a polar geodesic parametrization, E = 1, F = 0, G > 0. 

Proof. Since the w-parameter curves are unit-speed geodesies, we have 

E = x M »x M = 1, and x uu = 0. 

Thus 

P u == \X U * Ti. v ) u X u * X. vu jl u * JL UV ^-C'u O, 

so F is constant on each w-parameter curve. Now the functions E, F, G 
are well-defined even when x is not restricted to the strip S f . The v-param- 
eter curve v — > x(0, v) is simply the constant curve at the pole p, so 



Sec. 5] LENGTH-MINIMIZING PROPERTIES OF GEODESICS 343 

x v (0,v) = 0. 

But then F(0, v) = for all v, and since F u = 0, we conclude that F is 
identically zero. Because x (now restricted once more to the strip S f ) is a 
parametrization, that is a regular map, we know that EG — F 2 = EG is 
never zero. Hence G > 0. | 

5.5 Example We explicitly work out geodesic polar parametrizations 
in two classic cases. 

(1) The unit sphere 2 in E . For simplicity let p be the north pole (0, 0, 1 ). 
To get the geodesies radiating out from p as in Fig. 7.17, we change the 
geographical parametrization to 

x(u, v) = (sin u cos v, sin u sin v, cos u). 

Each w-parameter curve is indeed a unit-speed parametrization of a great 
circle, hence is geodesic. For u = we find 

x M (0, v) = (cos v, sin v, 0) 

= cos vei + sin ve 2 

where 

ei = C/i(p), e 2 = t/ 2 (p). 

Thus by the uniqueness of geodesies, 

X \U, V ) = 'Ycos vei + sin v ei \M ) 

which shows that x as denned above is the polar geodesic mapping (Defi- 
nition 5.2) . It is easy to see that the largest possible normal neighborhood 
9l e of p occurs when e = ir, for on the strip &, x is a polar parametrization 
of all the sphere except the north and south poles. 

(2) The hyperbolic plane H (Example 2.4). We choose p = (0, 0) and 




FIG. 7.17 



344 



RIEMANNIAN GEOMETRY 



[Chap. VII 



«i = Ui(p), 62 = £Mp)- (Since the function g is 1 at the origin, this is a 
frame.) We know from Example 4.11 that the geodesies of H through the 
origin follow Euclidean straight lines. Thus for any number v, the curve 

a{t) = (t cos v, t sin v) 

is at least a pre-geodesic of the type we are looking for. In Example 2.4 we 
found the arc length function s(t) = 2 tanh _1 (</2) for such a curve; thus 

s — > a I 2 tanh - 1 = ( 2 tanh - cos v, 2 tanh - sin v 1 

is the unit-speed reparametrization of a. Shifting notation from s to u, we 
obtain 



x(w, v) = I 2 tanh - cos v, 2 tanh - sin 



•) 



2 ' 2 

Since the w-parameter curves of x are unit-speed geodesies and 

x„(0, v) = cos vei + sin ve%, 

we conclude as in (1) that x is a geodesic polar mapping. The normal 
neighborhood in this case is the entire surface H. 

5.6 Theorem For each point q of a normal neighborhood 9l £ of p the 
radial geodesic segment in 9l t from p to q uniquely minimizes arc length. 

Proof. Let x be the polar parametrization of the normal neighborhood 
?fl f . If 

q = x(wo, v ), 

the radial geodesic segment is 

y(u) = x(u, v ), ^ u ^ Wo- 

Now let a be an arbitrary curve segment from p to q in M ; we may arrange 
for a to be denned on the same interval as y. 

We begin by proving 




L(y) ^ L(a). 



(1) 



FIG. 7.18 



First consider the case (Fig. 7.18) where 
a stays in the neighborhood 3l e . We may 
assume that once having left p, a never 
returns to p — if it did, throwing away 
this loop would only shorten a. Thus it 
is possible to write 

a(t) = xfa (0,02(0). 



Sec. 5] LENGTH-MINIMIZING PROPERTIES OF GEODESICS 345 

Since a(0) = p and a(u ) = q, we have 

ai(0) = ai(wo) = u 

(h(0) = v a 2 (w ) = v + 2im. 

(The term 2m results again from the nonuniqueness of angles in polar 
coordinates. ) 

Since for x we have E = 1 and F = 0, the speed of a is 

II a ' II = vV 2 + Ga 2 ' 2 . 



Now 
Hence 



Va/ 2 + Ga 2 ' 2 ^ \/V 2 = | ax | ^ a/ (3) 



<ft 



(4) 



L(a) = [ ° Voi' f + GV 2 d« ^ / a/ 
Jo J o 

= ax(uo) — ai(0) = Wo 

where the last step uses (2). But the radial geodesic y has unit speed, so 

/•«0 

L(y) = I dt = wo, 
Jo 

and we conclude that 

L(y) ^ L{a). 

If a does not stay in 9l e we have strict inequality, L{y) < L(a). For a 
must cross the poZar circle, u = w — indicated by a dashed line in Fig. 7.18 
— to escape from 9l«.f But by the proof above it must already have arc 
length at least u = L(y) when it reaches the circle. 

Now we prove the uniqueness assertion: 

If L(a) = L(y), then a is a reparametrization of 7 (5) 

The argument above shows that if L(a) = L(y), then a stays inside 
3fl« and the inequality in (4) becomes equality. The latter implies that 
V«i' 2 + Ga 2 ' 2 = th. Since G > we conclude from (3) that 

ax ^ 0, <U = 0. (6) 

Thus Oi has constant value v (so n = in (2 ) ) and 

a(t) = x(ai(0, v ) = y(a r (t)), 

showing that a is in fact a monotone reparametrization of 7. | 

t A careful proof will involve the Hausdorff axiom (Ex. 5 of IV.8) which we 
assume throughout this chapter. 



346 



RIEMANNIAN GEOMETRY 



[Chap. VII 




FIG. 7.19 



This fundamental result shows, as we remarked earlier, that if points 
p and q are close enough together, then — as in Euclidean space for arbitrary 
points — there is a unique geodesic segment from p to q which is shorter 
than any curve from p to q. (Unlike the Euclidean case however, there 
may be many other nowshortest geodesies from p to q.) If x is a geodesic 
polar parametrization at p we shall also call the route C ( of the v-parameter 
curve, u = e, the polar circle of radius e at p. (Fig. 7.19). Theorem 5.6 
shows that C t does in fact consist of all points at distance e from p. 

In special cases where large normal neighborhoods are available, this 
local information may be decisive. 

5.7 Example Length-minimizing properties of geodesies on the sphere 2 
of radius r. By a mere change of scale we can conclude from Example 5.5 
that each point p of 2 has normal neighborhood 9l, r : all of 2 except the 
point, — p, antipodal to the pole p. Hence Theorem 5.6 implies: 

(a) If two points p and q of 2 are not antipodal (that is, q 5* — p), 
then there is a unique shortest curve y from p to q. But we know all the 
geodesies of 2: 7 can only be the one that folllows the shorter arc of the 
great circle through p and q. 

(b) Intrinsic distance p on 2 is given by the formula 

p(p, q) = rd 

where t? (0 ^ t? ^ x) is the angle from p to q in 
E (Fig. 7.20). If p and q are not antipodal, this 
follows from (a), since 

P(p>q) = L(y) = re. 

As q moves toward the antipodal point — p of p 
we deduce by continuity that p(p, — p) = rir. 
Hence 




fig. 7.20 



Sec. 5] LENGTH-MINIMIZING PROPERTIES OF GEODESICS 347 




FIG. 7 21 

(c) There are infinitely many minimizing geodesies from a point p on 
2 to the antipodal point — p, namely (constant-speed parametrizations of) 
semicircles from p to —p. (Proof: These all have length rir = p(p, — p).) 

(d) No geodesic segment y of length L(y) > vr can minimize arc length 
between its end points. This follows immediately from the fact that in- 
trinsic distance p never exceeds ur. It is geometrically clear, since if y 
starts at p, its length exceeds irr as soon as 7 passes the antipodal point 
—p. But then the other arc 7* of the same great circle is shorter than 7. 

Suppose that a is a curve segment in M from p to q, and /3 is a curve 
segment from q to r. Now a and /3 cannot generally be united to form a 
single (differentiable) curve from p to r, since there may be a "corner" 
at q as in Fig. 7.21. Using the techniques of advanced calculus, one can 
"round off" this corner, obtaining a curve segment 7 from p to r which 
(to state the weakest theorem) is only slightly longer than a and /3. Ex- 
plicitly, for each e > there is a 7 such that L(y) ^ L(a) + L(0) + e. 

It follows that intrinsic distance satisfies the triangle inequality. In fact, 
given points p, q, and r the definition of intrinsic distance shows that for 
any e > there exist curves a and /3 as above such that 



LCIC CA1SL UU1VCS « £UU1 jJ na »UUVC OUl^ll uuau 

L(a) ^ p(p, q) + e, L(/3) ^ p(q, r) + 



e. 



Rounding off the corner at q costs at most another e: We get a curve 
segment 7 from p to r such that 

p(p, r) ^ L(y) ^ p(p, q) + p(q, r) + 3e. 

But since e is arbitrary, we conclude that 

p(p, r) ^ p(p, q) + p(q, r). 

5.8 Lemma If a is a shortest curve segment in M from p to q, then a 
is geodesic. 

Proof. We shall prove that if a : [a, b] — > M is a curve segment from p to 
q which is not a geodesic, then L(a) > p(p, q). But if a is not geodesic, 
then at some time £ the acceleration a" (to) is not zero. By continuity a." 
is nonzero near t , so we may assume t < b. For e > sufficiently small, 
at (to + e) lies in a normal neighborhood of a (to), and the segment of a from 



348 RIEMANNIAN GEOMETRY [Chap. VII 




FIG. 7.22 

to to to + e is not a geodesic, since a" {to) 9^ (Fig. 7.22). But then by 
Theorem 5.6 its length, L to ,t +t, is strictly greater than the intrinsic dis- 
tance from a (to) to a (to + e). Thus by the triangle inequality, 

L(a) = L a ,t + Lt ,t 0+( + Lt 0+f ,b 

> p(p, a(t )) + pM*o), a(t + e)) + p(a(* + e), q) 

^p(p,q). | 

This result is not too surprising: A shortest road can never turn. Nor 
will it have any corners, for a somewhat more sophisticated argument 
shows that a (possibly broken) shortest curve must in fact be an (un- 
broken) geodesic. 

We have now reached the main result of this section. 

5.9 Theorem Given any two points p and q in a geodesically complete 
geometric surface M , there is a shortest geodesic segment from p to q. 

Proof. The scheme is an ingenious one, worked out successively by several 
mathematicians. (See Theorem 10.9, p. 62, of Milnor [7].) We begin by 
selecting a candidate for shortest curve from p to q. Let 

P(v) = x(a, v), ^ v ^ 2ir, 

parametrize the polar circle C of radius a in a normal neighborhood of p. 
By Exercise 6 it follows that the function v — » p(/3(v), q) is continuous on 
the closed interval [0, 2w]; hence the function takes on its minimum value 
at say vo. Let 7 be the parameter curve, v = vo. Since M is geodesically 
complete, y(u) is defined for all u ^ 0. We will show that 7 hits q — in 
fact, that 

y(r) — q where r = p(p, q). (1) 

(This situation is illustrated in Fig. 7.23.) Since 7 has unit speed, it will 
follow that 

L( T ) == r = p(p, q), 

thus proving the theorem. 



Sec. 5] LENGTH-MINIMIZING PROPERTIES OF GEODESiCS 349 



q = y(r) 




FIG. 7.23 

To establish (1), we use a variant of the standard induction argument 
in which integers are replaced by real numbers. For each number u ^ 
consider the assertion 

a(u): p(7(w),q) = r - u (2) 

where, as above, r = p(p, q). This says that y (unit speed) is efficient: 
After traveling distance u, the distance to q has been reduced by precisely 
u. The proof vrill be finished ij we can prove that Gfc(r) is true, for then 

p(7(0,q) =0, 

so by Exercise 5, y (r) = q. We make a start on this by showing that &(a) 
is true for a as above; that is, 

p(7(«),q) = r - a. (3) 

According to Theorem 5.6, p(p, 7 (a) ) = a; hence by the triangle inequality 

r = p(p, q) ^ a + p(7(a), q). 

To get (3) we must reverse this inequality. By definition of intrinsic dis- 
tance, for any e > 0, there is a curve segment a from p to q such that 

L{a) ^ p(p,q) + e. 

Now a must hit the polar circle C, say at a(to), and we observe that the 
portion of a from p to a(t ) has length Li ^ a, and the remainder of a has 
length 

Lt ^ p(a(t ), q) ^ p(T(a),q). 

(The latter since 7(0) was a nearest point to q on C.) Thus 

a + p(7(fl), q) ^ Li + L 2 = L(a) ^ p(p, q) + e. 

Since e was arbitrary, we obtain the inequality 

a + p(y(«),q) ^ p(p, q) 
required to prove (3 ) . 



350 RIEMANNIAN GEOMETRY [Chap. VII 




P = 7(0) 

FIG. 7.24 

Now we turn to the inductive step of the proof. Since p cannot be nega- 
tive, Ct(w) is nonsense for u > r. Thus the set of numbers a for which 
(1(a) is true has a least upper bound b, with b ^ r. Since the functions 
involved in the assertion &(a) are continuous, it follows from the defini- 
tion of least upper bound that Ct(b) is true. 

Here is the plan of the rest of the proof: Assume b < r and deduce a con- 
tradiction. Then (since b ^ r) we must have b = r; hence Ct(r) is true, as 
required. 

Let C be a polar circle of radius a* < r — 6 in a normal neighborhood 
of y(b). By reproducing the argument for the circle C, we obtain a point 
c* such that 

p(c*,q) = p( 7 (6),q) -a*. (3') 

(See Fig. 7.24.) But a (6) reads p(y (6), q) = r - b, so 

p(c*, q) = r - b - a*. (4) 

The main step that remains is the proof of 

c* = y(b + a*). . (5) 

This is not too difficult. By the triangle inequality, 

p(p,c*) + p(c*, q) ^ p(p, q) = r. 
Using (4 ) we get 

p(p, c*) ^ b + a*. 

But there is a broken curve from p to c* whose length is precisely b + a*. 
In fact, referring to Fig. 7.24, we can travel on y from p to y (b) with arc 
length b, then from y(b) to c* on a radial geodesic with arc length a*. 
Thus by the remark preceding this theorem, this curve is not actually 
broken. Hence it is y all the way, so y(b -\- a*) is precisely c*. 
Finally, we substitute (5) in (4), obtaining 

p(y(b + a*), q) = r - (b + a*). 



Sec. 5] LENGTH-MINIMIZING PROPERTIES OF GEODESICS 351 

This says that &(b + a*) is true, and since b + a* is strictly larger than 
the upper bound b, we have the required contradiction. | 



EXERCISES 

1. In the hyperbolic plane H, find the intrinsic distance from the origin 
to an arbitrary point p. Deduce that all geodesies of H have infinite 
length, hence H is complete. (Hint: Use the triangle inequality.) 

2. For the Poincare half -plane (Exercise 6 of Section 4): 

(a) Find an equation F(x, y, c) = for the routes of the semicircular 
geodesies through the point (0, 1). 

(b) Find an equation G(x, y, a) = for the polar circles centered at 
(0, 1). (Hint: they are the orthogonal trajectories of the curves in 
(a).) 

(c) Make a sketch showing several curves from each family. 

3. At the point p = (r, 0, 0) of the cylinder M: x 2 + y 2 = r 2 , let 

d = (0, 1,0) and e 2 = (0,0, 1). 

Find an explicit formula for the mapping y (p. 341 ) in this case. What 
is the largest normal neighborhood of the point p? 

4. Test the scheme used to prove Theorem 5.9 in the special case M = E 2 . 
Explicitly: Starting from the geodesic polar mapping 

x(u, v) = (pi + u cos v, p 2 + u sin v) at p. 

follow the first paragraph of the proof of Theorem 5.9 to determine the 
geodesic y. 

5. Intrinsic distance is a metric on M. Show that 

(a) A normal neighborhood 9l« of p consists of all points q of M such 
that p(p, q) < e. 

(b) p satisfies the three metric properties: (i) p ^ 0; and p(p, q) = 
if and only if p = q, (ii) p(p, q) = p ( q , p ), and (iii) the triangle 
inequality. 

(Hint: It is necessary to use the Hausdorff axiom for the same 
purpose as in the footnote on page 345.) 

6. Intrinsic distance is continuous. In a geometric surface M, define p, — > p 
to mean that the sequence of real numbers p(p, p.) converges to 0. 
Prove that if p t - — > p and q t — ► q, then p(p t , q t ) converges to p(p, q). 

7. Let a and /S be two different unit-speed geodesies which start at the 
same point a(0) = 0(0). If a and meet again after having traveled 



352 RIEMANNIAN GEOMETRY [Chap. VII 

the same distance r > 0, that is, o(r) = P(r), prove that neither a or 
minimizes arc length past r. (Use the fact that broken geodesies 
cannot minimize arc length.) 

8. (Continuation). On the cylinder M: x + y 2 = r 2 , prove: 

(a) A geodesic starting at (a, b, c) cannot minimize arc length after it 
passes through the antipodal line t — * (—a, —b,t). 

(b) If q is not on the antipodal line of p, show that there is a unique 
shortest geodesic from p to q. 

Derive a formula for intrinsic distance on the cylinder. 

9. Show that the converse of Theorem 5.9 is false: Give an example of a 
geometric surface M such that any two points can be joined by a mini- 
mizing geodesic, but M is not geodesically complete. 

10. Let y: [a, b] — > M parametrize a portion of a meridian of a surface of 
revolution M. Prove that y uniquely minimizes arc length. (Hint- 
Express a competitive curve a as x(ai , a 2 ) where x is a canonical 
parametrization, and follow the scheme of Theorem 5.6.) 

11. Let M be an augmented surface of revolution (Ex. 12 of IV.l). 

(a) If M has only one intercept p (on the axis of revolution) show 
that every geodesic y of M starting at p uniquely minimizes arc 
length. 

(b) If M has a second intercept q show that the assertion in (a) holds 
if and only if y does not reach q. 

(Hint: No computations are needed.) 



6 Curvature and Conjugate Points 

We briefly examine the influence of Gaussian curvature K of a geometric 
surface M on the geodesies of M. 

6.1 Definition A geodesic segment y from p to q locally minimizes arc 
length from p to q provided that for any curve segment a from p to q 
which is sufficiently near 7 we have L(a) ^ L(y). 

To clarify the phrase "sufficiently near," we first define a to be e-close 
to y provided there is a reparametrization d of a, on the same interval / 
as 7, such that p(a(t), y(t)) < e for all t in I (Fig. 7.25). Then we change 
the ending of Definition 6.1 to "provided there exists an e > such that 
for any a which is e-close to 7, we have L(a) ^ L(y)." This local mini- 
mization is strict (or unique) provided we get strictly inequality 

L(a) > L(y) 
unless a is a reparametrization of 7. 



Sec. 6] CURVATURE AND CONJUGATE POINTS 353 




FIG. 7.25 

To get an intuitive picture of this definition we shall imagine that 7 is 
an elastic string — or rubber band — which (1) is constrained to lie in M; 
(2) is under tension; and (3) has its end points pinned down at p and q. 

Because 7 is a geodesic, it is in equilibrium: If it were not geodesic, its 
tension would pull it into a new shorter position. But is the equilibrium 
stable? If 7 is pulled aside slightly to a new curve a and released, will it 
return to its original position 7? Evidently 7 is (strictly) stable if and only 
if 7 is a (strict) local minimum in the sense above, for if a is longer than 7, 
its tension will pull it back to 7. 

Investigation of local minimization depends on the notion of conjugate 
points. If 7 is a unit-speed geodesic starting at p, then 7 is a w-parameter 
curve, v = v , of a geodesic polar mapping x with pole p. We know that 
along 7 the function G = x„« x„ is zero at u = 0, but is nonzero immediately 
thereafter (Lemma 5.4). A point 7 (s) = x(s, v ) with s > is a conjugate 
point 0/7 (0) = p on 7 provided G(s, v ) = 0. (Such points may or may not 
exist.) 

The geometric meaning of conjugacy rests on the interpretation of 
\/G = || x„ || as the rate at which the (radial geodesic) w-parameter curves 
are spreading apart. Roughly speaking, for fixed e > 0, if y/G = || x„ || 
is large, then the distance from x(w, v) to \(u, v + e) is large: The radial 
geodesies are spreading rapidly. When \/G is small, this distance is small, 
and the radial geodesies are pulling back together again. Thus when G 
vanishes at a conjugate point 

y(si) = x(si, v ), 

it suggests that for v near v , the u -parameter curves have all reached this 
same point after traveling (at unit speed) the same distance s x (Fig. 7.26). 
Unfortunately this meeting may not actually occur. (G controls only the first 
derivative terms, and higher-order terms may still be nonzero even though 
G vanishes.) 

The Euclidean plane E 2 should give the "standard" rate at which radial 



354 RIEMANNIAN GEOMETRY [Chap. VII 

ty(si) = x(si , » ) 



P = 7(0) 




FIG. 7.26 



geodesies spread apart, and for x(u, v) = (u cos v, u sin v), we have 

VG = u. 

In particular, there are no conjugate points. Let us compare the cases 
discussed in Example 5.5, the unit sphere S and hyperbolic plane H. For 
2, we find 

\/(j = sin u. 

Thus radial geodesies starting, say, at the north pole p of 2, spread less 
rapidly than in E 2 , since sin u < u for u > 0. Indeed one can see in 
Fig. 7.17 that after passing the equator they begin to crowd closer to- 
gether. All have their first conjugate point after traveling distance x, since 
•\/6r(x, v) = sin x = 0. In this case, of course, the meeting of geodesies 
actually takes place — at the south pole of 2. 

For the hyperbolic plane, we know that the geodesies radiating out from 
the origin are just Euclidean straight lines, but they are spreading more 
rapidly than in E 2 , as one may surmise from the fact that in H "rulers 
shrink as they approach the rim." To prove it we use the data in (2) of 
Example 5.5 to compute 

\/G = sinh u. 

Thus y/G > u for u > 0, and again there are no conjugate points. 

6.2 Theorem If 7 is a geodesic segment from p to q such that there are 
no conjugate points of p = t(0) on 7, then 7 locally minimizes arc length 
(strictly) from p to q. 

Proof. Let x be a geodesic polar mapping at p, and restrict its domain to 
the region in E 2 on which G > 0. Because there are no conjugate points of 
p on 7, we may write y(u) = x(u, v ) for ^ u ^ u . (Thus we allow 
u — in this equation, as usual, even though G = there.) Then let a 



Sec. 6] 



CURVATURE AND CONJUGATE POINTS 



355 




FIG. 7.27 



be another curve segment from p to q with a also defined on the interval 
[0, Wo]. Now our proof rests on the fact that if a is sufficiently close to y 
(as denned earlier), then a has an expression 

a(t) = x(ai(0, aj(0) 

which is so close to that of 7 that 

ai(0) = 0, / ai(wo) = u . 

(Fig. 7.27). 

A complete proof of this rather plausible assertion is not trivial. There 
is no trouble at u = 0, since we can replace a short initial segment of a 
by a radial geodesic — with no loss of generality, since this will not lengthen 
a. Then a x and a-t are constructed in step-by -step fashion using the fact 
that x is a regular mapping, and hence is locally a diffeomorphism. 

Then exactly as in the proof of Theorem 5.6 we have 

L(a) = / Vai' 2 + Ga 2 ' 2 dt ^ / 0/ dt 
Jo J o 

= ai(uo) — ai(0) = uo = L(y) 

and if L(a) = L(y), a is merely a reparametrization of 7. | 

Our task now is to free the notion of conjugate point from dependence 
on geodesic polar mappings. To do so we examine the "spreading coeffi- 
cient" y/G more closely. 

6.3 Theorem Let x be a geodesic polar mapping defined on a region 
where G > 0. Then -y/G = || x„ || satisfies the Jacobi differential equation 

(Vg) uu + kVg = 

subject to the initial conditions 

VG(0, v) = (VG)u(O, v) = 1 for all v. 
The restriction G > is needed to ensure that s/G is differentiate. 



356 RIEMANNIAN GEOMETRY [Chap. VII 

Now \/G(u, v) is actually well-defined for u = 0; indeed, 

VG(0,v) = ||x,(0,»)|| =0. 

However, y/G need not be differentiable at u = 0, so we shall interpret 
("\/G)u(0, v) and (\ZG) uu (0, v) as limits, for example, 

(VG)«(0, v) = lim„^ (y/G) u (u, v). 

Proof. The Jacobi equation follows immediately from Lemma 6.3 of 
Chapter VI, since as shown in Lemma 5.4, E = 1 and F = f or x. Thus 
by the remarks above, it suffices to prove 

lim (VG)u(u, v) = 1 (u > 0). 

u-»0 

We need only consider a single radial geodesic y(u) = x(w, v ), setting 

g(u) — \/G(u, v ) for u > 0. 

Again, since 1? = 1 and F = 0, we obtain a frame field 

Ei = 7' = x„, 1?2 = x„/flf on 7 for u > 0. 

Since 7 is a geodesic, E\ is parallel and by Exercise 3 of Section 4, so is 
Ei. By parallelism, Ei is thus well-defined at u = (Fig. 7.28). Now 

Ei(0) = x„(0, v ) = cos v ei + sin v e 2 
hence 

J5J 2 (0) = —sin vo ei + cos v e 2 . 
Furthermore, since E 2 is parallel and x„ = gEi on 7, we get 
x„„ = x vu = g E 2 on 7 for u > 0. 



^(0) 




Sec. 6] CURVATURE AND CONJUGATE POINTS 357 

Taking limits as u — > yields 

x„„ (0, v ) = (lim g (u) ) E 2 (0 ) . 



But 



hence 



x« (0, v) = cos i> ei + sin v e 2 for all v; 



x u »(0, y ) = —sin v ei + cos Voe 2 = Ei(0). 
Thus the last equation implies lim u -» </(0) = 1; that is, 

lim (y/G) u (u, v ) = 1 

u-»0 

for arbitrary v . | 

In terms of the spreading apart of radial geodesies, the initial conditions 
above show that as they first leave the pole p in any geometric surface, 
they are spreading at the same rate as in the Euclidean plane E 2 . For there 
we found \/G = u; hence 

VG (0, v) = 0, (VG)u(0, v) = 1. 

However, the Jacobi equation, written (\/G ! )«« = —K -\/G, shows that 
thereafter the rate of spreading depends on Gaussian curvature. For K < 0, 
radial geodesies spread apart faster than in E 2 . (We observed this earlier 
in the hyperbolic plane.) For K > the rate of spreading is less than in 
E 2 (as on the sphere). 

In particular, to locate conjugate points, it is no longer necessary to 
explicitly construct geodesic polar mappings, as we have done heretofore. 
We can find y/G on a geodesic 7 by simply solving the Jacobi equation on 
7, subject to the given initial conditions. Explicitly, Theorem 6.3 implies 
the following result. 

6.4 Corollary Let 7 be a unit-speed geodesic starting at the point p 
in M . Let g be the unique solution of the Jacobi equation on 7, 

g" + K(y)g = 

such that gr (0) = 0, </(0) = 1. Then the first conjugate point of 7(0) = p 
on 7 (if it exists) is 7 (si), where Si is the smallest positive number such that 
g(s t ) = 0. 

6.5 Example Conjugate points 

(1 ) Let 7 be a unit-speed geodesic starting at any point p of the sphere 
S of radius r. The Jacobi equation for 7 is thus g" + g/r 2 = 0, which has 
the general solution 



358 



RIEMANNIAN GEOMETRY 



[Chap. VII 



g(s) = A sin - + B cos - . 
r r 

The initial conditions g(0) = 0, ^'(O) = 1, then give g(s) = r sin (s/r). 
The first zero of this function with Si > occurs at Si = irr. Thus the first 
conjugate point 0/7(0) = p on 7 is at the antipodal point of p. (This agrees 
with our earlier computation for the unit sphere by means of geodesic 
polar mappings. ) 

(2) Let 7 be a unit-speed parametrization of the outer equator of a 
torus of revolution T with radii R > r > 0. Now 7 is a geodesic and on 7 
we know that K has the constant value l/r(R + r). Thus by Corollary 
6.4, the first conjugate point y(si) of 7(0) = p on 7 will occur at exactly 
the same distance s x along 7 as if 7 were on a sphere with this curvature 
K. It follows that si = 7r \/r(R + r). 

6.6 Corollary There are no conjugate points on any geodesic in a surface 
with curvature K ^ 0. Hence every geodesic segment on such a surface 
is locally minimizing. 

Proof. Apply Corollary 6.4 to a geodesic 7 in M. Since gr(0) = and 
g (0) = 1, we have g(s) ^ for s ^ 0, at least up to the first conjugate 
point (if it exists). But K ^ implies g" = — Kg ^ 0, so g is an increasing 
function; in fact g ^ 1. Hence g(s) ^ s up to the first conjugate point— 
which can thus never occur. The final assertion then follows from Theorem 
6.2. I 

For example, on a circular cylinder C (K = 0) the helical geodesic 7 
from p to q indicated in Fig. 7.29 is indeed stable, as one can verify by direct 
experiment. Although locally minimizing, it is certainly not minimizing. 
Evidently the straight-line segment a provides a considerably shorter 
way to get from p to q. 





FIG. 7.29 



FIG. 7.30 



Sec. 6] CURVATURE AND CONJUGATE POINTS 359 

To cany the study of conjugate points much further, it is necessary to 
use the calculus of variations (see Milnor [7]). We shall quote just one 
result, which supplements Theorem 6.2. As soon as a geodesic y starting at 
p passes the first conjugate point of p ony, it no longer locally minimizes arc 
length. This is fairly easy to see on a sphere 2. In Fig. 7.30 the geodesic y 
from p to q is only slightly longer than the first conjugate distance irr. 
If the plane of the great circle of y is rotated slightly about an axis through 
the end points p and q, it will slice from 2 a curve segment a which, as one 
can verify analytically, is strictly shorter than 7. (Note that the only 
shorter geodesic from p to q is not near y.) 

Theorem 6.3 can also be used to give a rather intuitive description of 
Gaussian curvature in an arbitrary geometric surface. 

6.7 Lemma If x is a geodesic polar mapping with pole p, then 

VG(u,v) =u- K(p) \ + o(u) (u ^ 0). 

6 

Throughout, o(u n ) denotes a function of u and v (u > 0) such that 
lim u _o o(u n )/u n = 0. In the formula then, if u is small enough, o(u ) is 
negligible compared to the first two terms. 

Proof. As before, consider g(u) = -y/G (u, v) on a radial geodesic 
7 (u) = x(u, v). As a solution of the Jacobi equation on 7, g is differentiate 
at u = 0. Thus it has a Taylor expansion 

g (u) = 0(0) + g(0)u + g"(0) ~ + </"(0) | + o(u 3 ). 

The initial conditions in Corollary 6.4 are g(0) = 0, gr'(0) = 1; hence from 
the Jacobi equation we get g" (0) = 0. Differentiating the Jacobi equation 
gives 

g" + K{y)'g + K(y)g' = 0. 

Hence 

9" (0) = -K(y(0)) = -K(p). 

Substitution in the Taylor expansion then gives the required result. | 

Suppose that the inhabitants of a geometric surface M want to deter- 
mine the Gaussian curvature of M at a point p. By measuring a short 
distance e in all directions fromp, they obtain the polar circle C e of radius e. 

Now if M = E 2 , the circumference of C e is just L(C E ) = 2ire. But for 
K > the radial geodesies from p are not spreading as rapidly, so C e 
should be shorter than 2xe; and for K < they are spreading more rapidly, 
so C t should be longer than 2x8. 



360 



RIEMANNIAN GEOMETRY 



[Chap. VII 




FIG. 7.31 

The relation between L(C e ) and K can be measured with some precision. 
For e > small enough, C e is parametrized by v -* x(e, v), where x is a 
geodesic polar parametrization at p. Thus 

L(Ce) = / VG (e, v) dv. 

Hence by the preceding lemma, 

L(C e ) = 2t (b - K(p) £ + o(e 3 )) . 



(*) 



Thus if surveyors in M measure L(C e ) very carefully for e small, they can 
determine approximately what the Gaussian curvature of M is at p. Taking 
limits yields 

6.8 Corollary K(p) = lim E ^ (3/7re 3 ) (2x8 - L(C e )). 

We can easily test formula (*) for a sphere S of radius r in E 3 . As in 
Fig. 7.31, the polar circle C e with center p is actually a Euclidean circle 
of Euclidean radius r sin &, where & = e/r. Thus by the Taylor series of 
the sine function, 

UC) = 2, (r si„ E) _ a, (e - ^ + o(e s )) 

which gives yet another proof that 2 has Gaussian curvature K = 1/r 2 . 



EXERCISES 

1. Let x be the polar parametrization of the hyperbolic plane given in 
Example 5.5. Derive s/G (m, v) = sinh u in two different ways: by 
computing x„ ° x v , and by solving the Jacobi equation. 



Sec. 6] CURVATURE AND CONJUGATE POINTS 361 

2. If C e is a polar circle around a point p of M, call the region enclosed by 
C e the polar disc D e of radius e. 
(a) Show that the area of the polar disc is 

A(D B ) =7r[e 2 -K(p)^ + o(e 4 )] 



hence 



w , 12 .. ire 2 - A(Z)e) 
J£(p) = — lim 



T e-*0 8* 

(b) Use this formula to find the Gaussian curvature of a sphere of 
radius r. 

3. At the origin in the hyperbolic plane, find the length of the polar 
circle C e and the area of the polar disc D e (0 < e < 2). Deduce from 
each result that K(0) = —1. 

4. Let Mbean augmented surface of revolution (Ex. 12 of IV.l). 

(a) If M crosses A at only one point p (as on a paraboloid of revolu- 
tion), show that p has no conjugate points on any geodesic. 

(b) If M crosses A at two points, p and q, (as on an ellipsoid of revolu- 
tion), show that p and q are conjugate on every geodesic joining 
them. (Hint: Theorem 6.3 of Chapter V provides a solution of the 
Jacobi equation.) 

The following exercises deal with a useful variant of the polar geodesic 
parametrization in which the pole p is replaced by an arbitrary regular 
curve. 

5. Let /3: / — » M be a regular curve in M, and let X be a (non vanishing) 
vector field on (8 such that /?' and X are linearly independent at each 
point. Define 

x(w, v) = yx(v)(u); 

thus the w-parameter curves of x are geodesies cutting across j8 with 
initial velocities given by X (Fig. 7.32). Prove: 

(a) x is a regular mapping on some region D containing the interval 
(0, v), v in /. 

(b) By a suitable choice of and X, this parametrization x becomes 
(i) the identity map of E 2 (natural coordinates), (ii) the canonical 
parametrization of a surface of revolution, and (iii) a ruled para- 
metrization of a ruled surface (Chapter V, Section 5). 

6. (Continuation). If is a umt-speed curve, and X is theunit normal JV 
of (Section 4), show that for x: E = 1, F = 0, and \^G is the solution 



362 



RIEMANNIAN GEOMETRY 



[Chap. VII 




FIG. 7.32 




, Vq), focal point 



FIG. 7.33 

of the Jacobi equation {y/G) uu + K\/G = such that 

VG (0, v) = 1 and (VG)u(O, v) = -k„(v). 

The natural choice of X in the preceding example means that G for this 
parametrization is geometrically significant. If G(it ,v ) = 0, then (by 
analogy with conjugate points) we say that x(u ,v ) is a focal point of 
along the normal geodesic v = v . Here light rays emerging orthogonally 
from |8 tend to meet (Fig 7.33). 

7. (a) If ,8 is a circle of latitude on a sphere S, show that the north and 
south poles of S are the only focal points of /3. 
(b) If fi is a curve in the Euclidean plane, show that its focal points 
are exactly its centers of curvature; that is, the points on itsevolute. 
(See Ex. 15 of II.4.) 



7 Mappings that Preserve Inner Products 

We have already seen that a local isometry F: M — > iV carries geodesies 
of M to geodesies of N. Using the notation y v for the geodesic with initial 
velocity v, we can be more explicit. 

7.1 Lemma If F: M — ► N is a local isometry, and v is a tangent vector 
to M at p, then 



Sec. 7] MAPPINGS THAT PRESERVE INNER PRODUCTS 363 

Proof. By the remark above, y = F(y v ) is a geodesic of N. Its initial 
velocity is the tangent vector 

t'(0) =F*( 7 /(0)) =^*(v) 

to N at F(p). Thus by the uniqueness of geodesies (Theorem 4.3), y is 

precisely 7f*(«)- I 

It follows that a local isometry is completely determined by its effect 

on just one frame. 

7.2 Theorem Let F and G be local isometries from M to N. If for some 
one frame ei, e 2 at a point p of M we have 

F*M = G*(ei), F*(e 2 ) = #*(«*), 
then F = G. 

Proof. If M is geodesically complete, the proof is particularly easy. If 
q is an arbitrary point of M, then by Theorem 5.9 there is a vector v at 
the special point p such that y v (r) = q. From the hypothesis on F* and 
G* , we deduce by linearity that F# and G* agree on v = Cid + c 2 e 2 . Thus 
the preceding lemma shows that 

F(y v ) = 7**<») = 7o*(«) = #(7«)- 
Hence, in particular, 

F(q) = F(y v (r)) = G(y v (r)) = G(q) 

or all points q of Af . 

The proof for M arbitrary is a refinement. Using Lemma 5.3 it is possible 
to get a broken geodesic from p to q and deduce F (/3) = 0(/8) by applying 
the argument above to each unbroken segment of 0. | 

We shall now use the fact that local isometries preserve geodesies to 
construct some local isometries. The goal is to demonstrate a family re- 
semblance among geometric surfaces of the same constant curvature. 
Given any number K, there is a particularly simple geometric surface 
M(K) whose Gaussian curvature has constant value K. 

(1) If K > 0, let M (K) be the sphere S of curvature K (hence radius 

i/Vk). 

(2) If K = 0, let M(K) be the Euclidean plane E 2 . 

(3) If K < 0, let M(K) be the hyperbolic plane H of curvature K 
(hence pseudo-radius \/y/ — K: see Exercise 4 of Section 2). 

We shall call M(K) the standard geometric surface of constant curvature 
K. Of course, there are many other constant curvature surfaces; these are 
distinguished by the fact that they are geodesically complete and simply 
connected (p. 176). 



364 RIEMANNIAN GEOMETRY [Chap. VII 

7.3 Theorem Let N be a geodesically complete geometric surface with 
constant Gaussian curvature K. Then there is a local isometry F of the 
standard surface M(K) onto N. 

The first mapping in Example 4.6 of Chapter VI is an instance of this 
theorem, as is the local isometry (Exercise 6 of Section 2) of a sphere onto 
a projective plane. 

Proof. The Case K < 0. We use the language of Example 2.4, where 
K = — 1. A mere change of scale (Exercise 4 of Section 2) takes care of arbi- 
trary K < 0. As in (2) of Example 5.5, let p be the origin of H = M (- 1 ), 
with ei = £A(p) and e 2 = £Mp). Let £i, e 2 be a frame at an arbitrary 
point of N. Then let x and x be the resulting geodesic polar mappings of 
H and N. 

For the surface N, we assert that 

(1) x is defined on the entire right half -plane S: u > (a consequence 
of geodesic completeness). 

(2) Its image x(S) covers all of N except possibly the pole p (a conse- 
quence of Theorem 5.9 and the definition of geodesic polar mappings). 

(3) x: S — > N is a regular mapping. (By Lemma 5.4, E = 1, and F = 0, 
but we have observed earlier that the Jacobi equation for K = — 1 gives 
VG = sinh u, so EG - F 2 = sinh 2 w > on S.) 

This general result is thus valid for x: S — » H as well, but here we know 
more. By Example 5.5 the whole surface H is a normal neighborhood of 
the pole p; thus x has only the usual ambiguities of polar coordinates; the 
equation x(u, v) = q determines u uniquely, and v uniquely but for addi- 
tion of some multiple of 27r (q ^ p). From this extra information we con- 
clude that the formula 

F(x(u, v)) = x(u, v) 

is consistent and thus defines a mapping F of H onto all of N. 

(To prove differentiability of F at the pole p we must resort, as in the 
proof of Lemma 5.3, to the mappings y and y corresponding to x and x.) 
It is easy to show that F is a local isometry by using the criterion of Lemma 
4.5 of Chapter VI. Indeed, from (3) above, we have 

E = 1 = E, F = = F, G = sinh 2 w = G, for u > 0, 

and at the pole p the preservation of inner products is an honest consequence 
of continuity. 

The Case K = 0. This is a word-for-word copy of the preceding argument 
except that 

M(K) = E 2 and G = G = u. 



Sec. 7] MAPPINGS THAT PRESERVE INNER PRODUCTS 365 

The Case K > 0. Here a new idea is required, since the largest normal 
neighborhood 91 of a point p in the sphere 2 = M(K) is not all of 2: The 
antipodal point — p is omitted. 

Arguing as in the case K < 0, we get a local isometry Fi: 91 — > N. Now 
repeat this argument once more at a point p* in 2 different from both p 
and —p. We obtain another local isometry Fz: 91* — > N where 91* is all 
of 2 except — p*. The frames determining F 2 are chosen so that the deriva- 
tive maps of Fi and F 2 agree at p*. Thus by Theorem 7.2, F\ and F 2 are 
identical on the overlap of 91 and 91*. But 91 and 91* cover the whole sphere 
2, so taken together F x and F 2 constitute a single local isometry F: 2 — ► N. 
Because 2 is compact and N is connected, Exercise 6 of Section 7 in Chapter 
IV shows that F carries 2 onto M. | 

An isometry F: M —> M of a geometric surface onto itself may be viewed 
as a symmetry of M . Every feature of the geometry of M is the same at each 
point p as at F(p), since this geometry consists of isometric invariants. 
The results of Exercise 9 of Section 4, Chapter VI, show at once that the 
set ${M ) of all isometries F: M —> M forms a group, just as do the set of 
all isometries of Euclidean space (Exercise 7 of Section 1 of Chapter III). 
We call d(M) the isometry group of M. 

This group d{M) is, of course, intrinsic to M, and when M is a surface 
in E 3 , should not be confused with the group S(M) of Euclidean symmetries 
of M (Exercise 7 of Section 8, Chapter VI). A Euclidean symmetry F of 
M C E 3 is an isometry of E 3 such that ¥{M) = M; these exist when the 
shape of M in E is symmetric in the ordinary sense of the word. Each 
Euclidean symmetry F of M gives rise to an isometry F | M: M — > M, but 
in general this process does not give all isometries of M C E 3 (Exercise 9). 

For an arbitrary geometric surface M, the isometry group &(M) gives 
a novel algebraic description of M. Roughly speaking, the more sym- 
metrical M is the larger d(M ) is. For example, although we shall not carry 
out the proof, the ellipsoid 

M: W> + t + -t = l (a>b> c) 
a z b z c l 

has exactly eight elements in its isometry group, these all arising from its 
Euclidean symmetries as described above: three reflections (one in each 
coordinate plane), three 180° rotations (one around each coordinate axis), 
the isometry p — ► — p, and, of course, the identity map of M . 

The smallest possible isometry group d(M) occurs when the identity 
map of M is the only isometry of M. We can produce such a geometric 
surface by putting a bump on the ellipsoid in such a way as to destroy all 
seven of its nontrivial isometries. 

By contrast, a geometric surface M has the maximum possible sym- 



366 



RIEMANNIAN GEOMETRY 



[Chap. VII 




metry when every possible isometry permitted by Theorem 7.2 actually 
exists. That is, given frames ei, e 2 and ei , e 2 at any two points of M , there 
exists an isometry F: M —*■ M such that 

F*(e x ) = ei, F*(e 2 ) = e 2 . 

In this case, we shall say that M is frame-homogeneous; any two frames on 
M are symmetrically positioned. 

Thus what we proved in Theorem 2.3 of Chapter III is that E 3 is frame- 
homogeneous, and the same proof is valid for arbitrary E ra , in particular 
for E . In the exercises for this section, we shall see that every standard 
surface M(K) of constant curvature is frame-homogeneous. 

7.4 Definition A geometric surface M is point-homogeneous (or merely 
homogeneous) provided that given any two points p and q of M there is an 
isometry F: M — > M such that F(p) == q. 

A frame-homogeneous surface, is of course, homogeneous — but not 
conversely. A circular cylinder C in E furnishes an example. In fact, if 
F is a rotation of E 3 about the axis of C, or a translation of E 3 along this 
axis, then F carries C onto C, producing an isometry of C. Hence given any 
points p and q of C, we can first rotate to bring p to p on the same ruling 
as q, then translate p to q. The composition of these two isometries is an 
isometry carrying p to q. On the other hand, C is not frame-homogeneous: 
all its points are geometrically equivalent, but not all its frames. (Proof: 
For the unit vectors shown in Fig. 7.34, no isometry could carry ei to ei, 
for by Lemma 7.1, F would have to send the one-to-one geodesic y e to 
the periodic geodesic 7^ — an impossibility, since F is one-to-one.) 

Homogeneity is a very strong restriction. 

7.5 Theorem If a geometric surface M is homogeneous, then M is 
geodesically complete and has constant Gaussian curvature. 

Proof. Constancy of curvature follows immediately from the definition 
of homogeneity and the fact that isometries preserve curvature. The proof 



Sec. 7] MAPPINGS THAT PRESERVE INNER PRODUCTS 367 

F*(u) = a'iU) 





FIG. 7.35 

of completeness is more interesting. If M is not geodesically complete, there 
is a maximal unit-speed geodesic a defined only on an interval, say, /: 
t < a, which is not the whole real line. Let us show that this is impossible. 
By Lemma 5.3, all geodesies emanating from some arbitrary point p of 
M run at least for some fixed distance e > 0. Choose t in I such that 
a — to < e/2. Because M is homogeneous, there is an isometry F: M — *• M 
such thatF(p) = a(t ). Now for some unit vector u at p, F* (u) = a {to). 
Thus the geodesic segment F(y u ) has initial velocity 

F*(yJ(0)) = F*(u) = a (k) 

and runs for distance e at unit speed (Fig. 7.35). But then a shift of para- 
metrization enables us to apply Theorem 4.3 and thereby define a on the 
interval I*: t < t + e. But t + e > a, so this contradicts the maximality 
of the interval /, and thus proves M is geodesically complete. | 

As the title of this section suggests, (local) isometries are not the only 
inner-product-preserving mappings of importance in geometry. We shall 
take a brief look at the other main types. 

7.6 Definition Let F: M — > E 3 be a mapping of a geometric surface into 
E . If the derivative map F* preserves inner products of tangent vectors, 
then F is an isometric immersion. If F is also one-to-one, then F is an 
isometric imbedding. An isometric imbedding F such that the inverse func- 
tion F~~ : F(M) — > M is continuous is said to be proper. 

This definition is unduly restrictive. Evidently we could replace E 3 — or 
even M — by any Riemannian manifold (p. 308). 

7.7 Lemma If F: M — * E 3 is a proper isometric imbedding of a geometric 
surface M into E 3 , then the image F(M) is a surface in E 3 and the function 
F: M — * F(M) is an isometry. 

Proof. If x: D — ► M is a proper patch in M, then the composite mapping 
F(x) : D — » E 3 is a patch that lies in F(M ). Furthermore, F(x) is a proper 
patch. In fact, its inverse function F(x(D)) —> D is just x -1 /^ -1 , which is 
continuous since x -2 and F' 1 are continuous. Thus we can easily check 



368 RIEMANNIAN GEOMETRY [Chap. VII 

Definition 1.2 of Chapter IV. Now as a geometric surface, F(M) uses the 
dot product of E , and by definition F: M — > E 3 preserves inner products. 
Hence when considered as a mapping of M onto F{M), F preserves inner 
products. | 

Thus the study of the geometry of surfaces in E 3 is exactly the same as 
the study of proper isometric imbedding s of geometric surfaces into E 3 . 
This rather technical fact is important only because it suggests a considera- 
ble generalization of our work in Chapters V and VI. We could just as well 
have studied the far larger class of isometric immersions into E 3 , dropping 
the one-to-one and properness restrictions. There is, in fact, no real diffi- 
culty involved, except for complications of notation. 

As in the special case discussed on page 308, the image F(M) of an 
isometric immersion F: M — > E 3 may cut across itself; nevertheless, we 
shall think of it as a kind of defective surface in E 3 . If we define the shape 
operator of such an immersed surface, this should at least suggest how to 
generalize the rest of Chapters V and VI. 

Because inner products are preserved, an isometric immersion F is 
regular. Thus F*(T P (M)) is a two-dimensional subspace of 7V( P )(E 3 ); it 
plays the role of a tangent plane for F(M) at F(p). A unit normal function 
U assigns to each point p (in some region of M ) a unit vector orthogonal 
to F* (T P (M)). If a is a curve in M, then U a is a vector field on F(a) in 
E 3 . Then if v is the initial velocity of a, we define S(\) to be the unique 
vector in T P (M) such that 

F*(S(v)) = - Uj(0) 

This shape operator S is again a symmetric linear operator on T P (M). 

Most of our earlier results hold up rather well under generalization. 
For example, if the Gaussian curvature K of M is defined intrinsically as 
in Section 2 — then by reorganizing the logic in Chapter VI, Section 2, we 
can show that K — det S. 

A theorem such as Theorem 3.7 of Chapter VI becomes more informative: 
If M is a compact surface with constant curvature K(>0), and F: M —*■ E 3 
is an isometric immersion, then F is an isometry of M onto a Euclidean 
sphere 2 of radius 1/s/K in E . 

In other words, even if we give F(M) permission to cut across itself, 
this cannot happen: F{M) can only be an ordinary round sphere in E . 

We have seen that there are geometric surfaces M which cannot be 
isometrically imbedded in E 3 — for example, the flat torus (Example 2.5) 
or the projective plane, Exercise 6 of Section 2. In this case it is natural 
to try to imbed M in a higher dimensional Euclidean space E n . The larger 
n is, the less difficult the task becomes. (Roughly speaking, with more 
dimensions available for M to curve through, there is a better chance that 



Sec. 7] MAPPINGS THAT PRESERVE INNER PRODUCTS 369 

a shape can be found for M which is compatible with its intrinsic geometry. 
See Chapter VI, Section 9.) 

Thus, although there are no flat toruses in E 3 , they can be found in E 4 . 

7.8 Example An isometric imbedding of a flat torus in E 4 . Start with 
the mapping x: E 2 — > E 4 such that 

x(u, v) — (cos u, sin u, cos v, sin v). 

If x is the parametrization of the flat torus T given in Example 2.5, then 
the formula 

F(x(u, v)) = x(u, v) 

is consistent; in fact it defines a one-to-one mapping F: T — > E 4 . The proof 
consists in observing that 

x(u, v) = x(wi, vi) <=>t*i = u + 27rra,i>i = v + 2ira <=> \(u, v) = x(u u t> x ). 

Reading the implication arrows from left to right we get the required 
consistency; the reverse direction shows that F is one-to-one. 

Then F is an isometric imbedding provided F* preserves inner products. 
In the usual fashion we compute 

x u = ( — sin u, cos v, 0, 0) 

x„ = (0, 0, —sin v, cos v) 
Hence 

E = l, F = 0, (5=1. 

These functions agree with E, F, and G for x, so exactly the same argument 
used to prove Lemma 4.5 of Chapter VI shows that F * preserves inner 
products. 

The general situation here is not well understood. Although every 
compact geometric surface can be isometrically imbedded in E , it re- 
mains a possibility that 17 can be replaced by as low a dimension as 4. 



EXERCISES 

1. Let F: M — ► N be a local isometry, and suppose that M is geodesically 
complete. Show that F is onto if and only if N is geodesically complete.! 

2. Prove that a geodesically complete geometric surface with constant pos- 

t Though the proof is not elementary, it is known that both of these properties 
are consequences of the geodesic completeness of M. 



370 RIEMANNIAN GEOMETR\ [Chap. VII 

itive curvature is compact. (The result still holds if merely K ^ c > 0. 
See Myers' theorem in Hicks [5].) 

3. Suppose that in M any two points can be joined by at least one geodesic, 
and that in N any two points can be joined by at most one geodesic. 
Prove that every local isometry F: M — * N of such surfaces is one-to- 
one. 

4. Let F: M — » M be an isometry that is not the identity mapping. If a 
unit speed curve is fixed under F, that is, 

F(a(s)) = a(s) for alls, 

show that a is a geodesic of M. 

5. Let x and x be geodesic polar parametrizations of normal neighbor- 
hoods 9l E and 9l E (same e) in two geometric surfaces. If K(x) = K(x) 
on the common domain S e of x and x, prove that 9l e and 3l E are iso- 
metric. 

6. Prove that the sphere S and hyperbolic plane H are frame homogene- 
ous. (Hint: for 2 derive the required isometries from orthogonal 
transformations of E 3 ; fori/ use Theorem 7.3 and a preceding exercise.) 

7. Show that the flat torus (Example 2.5) is homogeneous, but not frame 
homogeneous, and that an ordinary torus of revolution in E 3 is not 
homogeneous. 

8. Prove: 

(a) For the right circular cylinder C: x 2 + y 2 = r 2 in E 3 , every isometry 
F: C -> C has the form 

^(p) = (Pi cos & ± p 2 sin #, p x sin # ± p 2 cos t?,ep 3 + a) 

where e = ±1. 

(b) Every isometry of a sphere or right circular cylinder in E 3 is the 
restriction of an isometry of E 3 . 

9. Let M be the cylinder in E 3 whose cross-sectional curve is the ellipse 
4a; + y 2 = 4. (Any other closed noncircular curve could be used.) 
Show that there is an isometry of M which is not the restriction of an 
isometry of E 3 . (Hint: parametrize M by x(u, v) = a(u) + vU 3 , 
where a is a periodic unit-speed parametrization of the ellipse.) 

10. In the sphere 2 of radius r, let T be a triangle whose sides are geodesic 
segments of lengths a, b, and c (all less than vr). Let # be the angle 
of T at the vertex p opposite side a. 
(a) Prove the law of cosines: 

a b c , . b . c 

cos - = cos - cos - + sin - sin - cos #. 



Sec. 7] MAPPINGS THAT PRESERVE INNER PRODUCTS 371 

(b) Show that this formula approximates the usual Euclidean law 
of cosines when r is large compared to a, b, c. 

{Hint: to determine cos # find unit vectors 115, u c at p tangent 
to the sides b and c. ) 

11. Prove that the projective plane (Exercise 6 of Section 2) is frame 
homogeneous. (Hint: if F : 2 — > 2 is an isometry of the sphere 2 C E 3 , 
then F( — p) = — ^(p), hence there is a mapping F: 2 — > 2 such that 
PF = FP.) 

12. Show that the isometry groups of isometric surfaces are isomorphic. 

13. If M is a surface in E 3 that does not lie in a plane, show that the func- 
tion F —> F | M is an isomorphism of the Euclidean symmetry group 
S(M) onto a subgroup of the isometry group d(M). 

Isometries of the hyperbolic plane may be constructed explicitly by 
recognizing a point of the plane as a complex number 

z = u + iv = (u, v), 
and using exercises of Section 1. Thus if | z | denotes the magnitude of z, 

I I 2 - 2,2 

\z\ = zz = u + v , 

the hyperbolic plane may be described as the disc | z \ < 2 with conformal 
geometric structure given as in Example 1.3 by g(z) = 1 — \z\ 2 /4. 

1 4. (A translation of the hyperbolic plane. ) For a fixed real number c = (c, ) 
in H, let T be the mapping T(z) = 4 [(z + c)/(cz + 4)] defined on H. 

(a) Show that T(H) <z H, and that T: H — > H is one-to-one and onto. 
If H denotes the same disc, \z\ < 2, but with the usual Euclidean 

structure, then Exercise 7 of Section 1 shows that T: H' — ► H' is a 
conformal mapping with scale factor X (z) = | dT/dz | . 

(b) Show that this scale factor is 



X(z) = 4 



4 - c z 

cz + 4 | 2 



(c) Deduce that T: H — > H is an isometry of the hyperbolic plane. 
(Hint: Use Ex. 9 of VII. 1.) 

These methods can be used to show that H is frame-homogeneous, 
and — carried somewhat further — to give an elegant derivation of the 
geodesies of H. 

15. (The Poincare half -plane P is isometric to the hyperbolic plane H.) 
In terms of complex numbers, P is the half-plane Im z > with con- 
formal geometric structure g(z) = Im z. (Im z is the imaginary part 
v of z = u + iv. ) Let F be the mapping 



372 RIEMANNIAN GEOMETRY [Chap. VII 

F(z) = Z + 2i 
{Z) iz + 2> 

defined on H. Show that 

(a) Im F(z) = (4- \ z\ 2 )/ \iz + 2 \ 2 . 

(b) F is a one-to-one mapping of H onto P. (Compute F _1 explicitly.) 

(c ) Relative to Euclidean structures, F is conformal, with scale factor 
X (z) = 4/ | iz + 2 | 2 . 

(d) F: i/ — ► P is an isometry. 

Make a sketch of i7 and P indicating the images in P of each of the 
four quadrants of H. 



8 The Gauss-Bonnet Theorem 

We have seen that the Gaussian curvature K of a geometric surface M 
has a strong influence on other geometric features of M such as parallel 
translation, geodesies, isometries, and, of course, the shape of M if it 
happens to be in E . Now we will show that the influence of Gaussian curva- 
ture penetrates to the ultimate topological conformation of M — to proper- 
ties completely independent of the particular geometric structure on M. 

The main step in this proof is a theorem which relates the total curva- 
ture of a 2-segment to the total amount that its boundary curve turns. 

For an arbitrary curve a in M, the geodesic curvature tells its rate of 
turning relative to arc length. Thus to find the total amount a turns, we 
integrate with respect to arc length : 

8.1 Definition Let a: [a, b] — > M be a regular curve segment in an oriented 
geometric surface M. The total geodesic curvature /« k ds of a is 

.*(b) 

K g (s) ds 



•'s(a) 

where k (s) is the geodesic curvature of a unit-speed reparametrization of 
a. 

The total geodesic curvature of a in M is thus an analogue of the total 
Gaussian curvature of a surface in E 3 . For example, let C be a circle of 
radius r in E 2 , where E 2 has its natural orientation. If a is a curve making 
one counterclockwise trip around C, then a has constant geodesic curvature 
Kg = l/r. Thus 

/ Kg ds = - 2-n-r = 2t, 
K r 

regardless of the size of the circle. A clockwise trip around C will have total 



Sec. 8] THE GAUSS-BONNET THEOREM 373 

curvature — 2ir, for in general: If the orientation of M is kept fixed, then 
the total geodesic curvature of a curve segment a is unaffected by orien- 
t&tioii-preserving reparametrization, but has its sign changed by an orien- 
t&tion-reversing reparametrization. (The former is clear from the defini- 
tion; the latter can be deduced, for example, from the following lemma.) 

8.2 Lemma Let a: [a, b] — *■ M be a regular curve segment in a region of 
M oriented by a frame field E u E%. Then 



J Kg ds = <p(b) — <p(a) + J 



«12 



where <p is an angle function from Ei to a on a, and w^ is the connection 
form of E x , E% 

Proof. None of these terms are affected by orientation-preserving 
reparametrization; thus we may assume that a is a unit-speed curve. But 
then the result follows immediately by integrating the formula in Lemma 
4.5 | 

For the integration theory in Chapter VI, Section 7, we used 2-segments 
x: R — > M which were one-to-one and regular on the interior R° of R. 
Now we shall impose the more stringent requirement that x be one-to-one 
and regular on the boundary of R as well. (This is equivalent to saying 
that x: R — *• M is the restriction to R of a patch defined on some open set 
containing R.) 

When x is a one-to-one regular 2-segment, its edge curves a, 0, y, 8 
(Definition 6.4, Chapter IV) are one-to-one regular, and we shall think 
of the boundary dx = a + (i — 7 — 5 as a single broken curve enclosing 
the rectangular region x(R). We now want to define the total geodesic 
curvature of dx. The definition of geodesic curvature shows that the total 
geodesic curvature of a curve is simply the total angle that its unit tangent 
T turns (relative to arc length). But to go all the way around 

dx = a -\- & — 7 — 5 

we must get not merely the total turning on the edge curves; that is, 



/ Kg dS = I Kg dS -{- I Kg dS + I Kg rfS + / 

= / Kg dS + / Kg dS — f Kg dS — IK 
J a JB J-v J S 



Kb 
& 



ds 

s 



but also the angles through which a unit tangent T on dx would have to turn 
at the four corners of the rectangular region x(R) (Fig. 7.37). For 



R: a ^ u ^ b, c ^ v ^ d 



374 



RIEMANNIAN GEOMETRY 



[Chap. VII 




0(0) 



-«'(*) 



FIG. 7.36 




FIG. 7.37 



these "corners" 

pi = x(a,c), pa = x(b,c), p 3 = x(6,d), p 4 = x(a,d) 

are called the vertices of x(R). 

In general, if a regular curve segment a in an oriented region ends at the 
starting point of another segment /3, say a (I) = /8(0), then the turning 
angle e from a to /3 is the oriented angle from a (1 ) to /3' (0) which is smallest 
in absolute value (Fig. 7.36). For a 2-segment we use the orientation 
determined by x, that is, the area form dM such that dM(x u ,x v ) > 0, to 
establish some terminology which is familiar in the case of a polygon in 
the plane. 

8.3 Definition Let x: R — » M be a one-to-one regular 2-segment, with 
vertices pi, p2, p3, p 4 . The exterior angle ey of x at py (1 ^ j " ^ 4) is the 
turning angle at py derived from the edge curves a, 0, —7, —8, a, • • • in 
order of occurrence in dx. The interior angle ty of x at py is t — ey (Fig. 7.37). 

This definition is given with more general applications in mind; in the 
case at hand, exterior angles can easily be expressed in terms of the usual 
coordinate angle 1? from x„ to x„ (0 < & < it) by 

81 = X — #1 82 = #2 83 = 7T — #3 84 = t?4 

where #y is the coordinate angle at the vertex py. For example, let us con- 



Sec. 8] THE GAUSS-BONNET THEOREM 375 




FIG. 7.38 

sider the situation at p 3 , as shown in Fig. 7.38. By the definition of the 
edge curves, ($' is x„, but (—7)' is — x u , since —7 is an orientation-reversing 
reparametrization of 7. Thus e 3 + #3 = ir. (Analytical proofs may be 
based on the definition of oriented angle in Section 7, Chapter VI.) 
We are now ready to prove the fundamental result of this section. 

8.4 Theorem Let x: R — > M be a one-to-one regular 2-segment in a 
geometric surface M. If dM is the area form on x{R) determined by x, 
then 

ff KdM + f Kg ds + (ex + e 2 + e 3 + e 4 ) = 2* 

*• v ^ v N/ , . ^ 

total Gaussian total geodesic 

curvature of x curvature of 3x 

(The geodesic curvature and exterior angles use the orientation of x(^) 
given by dM, where dM (x„,x„) > 0. Note that M itself need not be ori- 
ented — or even orientable.) 

We call this result the Gauss-Bonnet formula with exterior angles. Since 
Bj = ir — ij for 1 ^ j '^ 4, the formula may be rewritten 

f[ KdM + f Kg ds = (n + i 2 + 13 + i 4 ) - 2tt 

in terms of the interior angles oix(R). 

Proof. Let Ei = x u /%/E on the region x(R). Then let E 2 be the unique 
vector field such that E h E 2 is a frame field with dM '(Ei, E 2 ) = +1. In 
this case (compare with p. 292), the second structural equation becomes 

dun = -Kdi a 2 = -K dM 

The power for this proof is supplied by Stokes' theorem (6.5 of Chapter 




376 RIEMANNIAN GEOMETRY [ Chap . V || 

IV) which gives 

// KdM + f co 12 = 0. (1) 

"^ " x « J Let us now use Lemma 8.2 to evaluate 

/ OJ12 = / CO12 + / 0)12 — / C012 — / C0l 2 . (2) 

FIG. 7.39 3x J « J * J y _ Js 

On a we have a = x u = y/E E u so the 
angle <p from E\ to a is identically zero. Thus by Lemma 8.2, 

/ wi2 = / K g ds. (3) 

J a J a 

Next we try a harder case, say J s o>i 2 . Here the angle <p from 

El = VE toS ' = Xv 

is precisely the coordinate angle # from x M to x v (See Fig. 7.39.) Hence by 
Lemma 8.2 we get 



/ k ds = & 4 — #i + f 



OJ12 



where, as above, < & 3 < t is the coordinate angle at the vertex p y of x 
(1 ^ j ^ 4). But since 

#i = ir — ei and #4 = 8 4 , 
This becomes 

/ W12 = t — 81 — 84 + / k<7 ds. (4) 

In an entirely similar way we find 

/ 0>12 = — IT + 8 2 + 83 + I Kg dS (5) 

and 

J cow = J K fl ds. (6) 

Thus (2) becomes 

/ «u = / Kg ds -\- I Kg ds — / Kg ds — I Kg ds — 2tt + (ei + e 2 + e 3 + e 4 ) 

•'ax •'a «/0 J T Jj 

= / k ds — 2x + (ei + e 2 + e 3 + 84). 
Substitution in (1) then yields the required formula. | 



Sec. 8] 



THE GAUSS-BONNET THEOREM 



377 





FIG. 7.40 



The Gauss-Bonnet formula actually depends not on the particular 
mapping x: R — » M, but only on its image (R = x(R). Explicitly, if x is 
another one-to-one regular 2-segment x with the same image (R, then each 
of the six terms in the Gauss-Bonnet formula for x has exactly the same 
numerical value as the corresponding term for x. This should be no sur- 
prise if x and x have the same orientation, that is, determine the same 
area form on (R. But suppose they have opposite orientation (as in Fig. 
7.40) so that dM x = —dM x . To take the trickiest case, consider correspond- 
ing edge curves such as a. and in the Fig. 7.40. Now a and /8 run in 
opposite directions: /8 is an orientation-reversing reparametrization of a. 
But the geodesic curvatures of a and j8 are computed in terms of the op- 
posite area forms dM x and dM x . Thus there are two sign changes, so 



/. Ka ^ = i 



ds. 



The remainder of this section will be devoted to applications of the Gauss- 
Bonnet formula. The basic idea is to extend it to more general regions — 
in particular to entire geometric surfaces. To do so we must look at some 
fundamental properties of surfaces which do not involve geometry. 

A rectangular decomposition 3D of a surface M is a finite collection of 
one-to-one regular 2-segments xi, • • • , x/ whose images cover M in such 
a way that if any two overlap they do so in either a single common vertex 
or a single common edge. 

Evidently a rectangular decomposition is a special kind of paving (Defi- 
nition 7.3 of Chapter VI), but the regions Xi(Ri) are now really "rectangu- 
lar" (since x» is one-to-one regular on all of Ri) and they are required to 
fit together very neatly, as in Fig. 7.41 (compare the paving in Fig. 6.17). 

8.5 Theorem Every compact surface M has a rectangular decomposition. 

(Hence in particular M has a paving.) This result is certainly plausible, 
for if M is made of paper, we could just take a pair of scissors and cut out 
rectangular pieces until all of M is gone. A general proof is given in Lef- 
schetz [8] (use Exercise 10). 

We shall understand that a rectangular decomposition 3D carries with it 
not only its rectangular regions Xi(Ri) — called faces — but also the vertices 
and edges of these regions. 



378 



RIEMANNIAN GEOMETRY 



[Chap. VII 




FIG. 7.41 

8.6 Theorem If 3D is a rectangular decomposition of a compact surface 
M , let v, e, and / be the number of vertices, edges, and faces in D. Then 
the integer v — e -+- / is the same for all rectangular decompositions of M . 
This integer x(M) is called the Euler-Poincare characteristic of M. 

The natural proof of this famous theorem is purely topological, however 
it is an easy consequence of Theorem 8.8. 

These results may easily be generalized. First, instead of an entire surface, 
we could deal with a polygonal region, one which can be decomposed into 
rectangular regions Xi(Ri) fitting together neatly (as above). Second, 
we could everywhere replace rectangles R by polygons. (A polygon P is the 
bounded region in E enclosed by a simple polygonal curve — P including 
this curve.) Combining both generalizations we get the concept of polyg- 
onal decomposition 3D of a (polygonal) region (R in M. The Euler-Poincare 
characteristic x(#0 of (R is still independent of the choice of polygonal 
decomposition 3D. 

8.7 Example Euler-Poincare characteristic. 

(1) A sphere 2 has x(2) = 2. By "inflating" a cube as in Fig. 7.42 we 
get a rectangular decomposition 3Di of 2. SDi has v = 8, e = 12, / = 6 — 
and thus x = 2. Inflating a prism gives a polygonal decomposition SD 2 
with v = 6, e = 9, / = 5— again x = 2 (Fig. 7.42). 




^l^ 



Sec. 8] THE GAUSS-BONNET THEOREM 379 





M + H = M' 
FIG. 7.43 

(2) A torus T has x(T) = 0. Picture T as a torus of revolution, and cut 
it along any three meridians and three parallels. This gives a rectangular 
decomposition 2D for which v = 9, e = 18, / = 9, hence x = 0. 

(3) Adding a handle to a compact surface reduces its Euler-Poincare* 
characteristic by 2. 

Roughly speaking, a "handle" is a torus with the interior of one face 
removed. (We suppose that M and the torus are given in some rectangular 
decomposition.) To add a handle to M, remove the interior of a face of M 
also, and to the resulting rim smoothly attach the rim of the handle, so 
that the vertices and edges of the two rims coincide (Fig. 7.43). 

This operation produces a new surface M already equipped with a rec- 
tangular decomposition. The Euler-Poincare" characteristic of M' is 

x(M) - 2, 

since its decomposition has exactly two faces less than M and the torus 
combined. (Coalescing the two rims eliminates four vertices and four 
edges as well, but this has no effect on x-) 

It is easy to see that diffeomorphic surfaces have the same Euler-Poincare 
characteristic, for if x x , • • • , x/ is a decomposition of M and F: M — > M 
is a diffeomorphism, then F(xi), • • • , F(x f ) is a decomposition of M with 
exactly the same v, e, and /. 

For example, no matter how wildly we distort the sphere 

2:z 2 + ?/ 2 + z 2 = 1, 

the resulting surface M will still have Euler-Poincare" characteristic 2. 
So long as no geometric structures are involved, the word "sphere" might 
be used to mean "a surface diffeomorphic to 2." To avoid any possible 
confusion we shall stay with the longer terminology. 

Suppose we start with the sphere 2 and successively add h handles 
(h = 0, 1, 2, • • • ) to obtain a new surface 2 (ft). What is remarkable about 
the operation of adding handles is that every compact orientable surface M 
is diffeomorphic to some 2 (ft). In this case we shall say that M itself has 
ft handles. By (3) of Example 8.7, it follows that 

x(M) = X (2(ft)) = x (2) - 2ft = 2 - 2ft. 



380 



RIEMANNIAN GEOMETRY 



[Chap. VII 




FIG. 7.44 

In Fig. 7.44, for example, all four surfaces have just one handle, and thus 
all have x = 0. 

Although we have used the concepts of calculus in this brief discussion 
of the Euler-Poincare" characteristic, our remarks remain valid if differen- 
tiability is everywhere replaced by continuity. The Euler-Poincare* char- 
acteristic is, in fact, a topological invariant.^ 

Returning now to the subject of geometric surfaces, we prove a spectacu- 
lar consequence of Theorem 8.4. 

8.8 Theorem (Gauss-Bonnet) If M is a compact orientable geometric 
surface, then the total Gaussian curvature of M is 2irx(M), where xiM) 
is the Euler-Poincare* characteristic of M. 

Proof. Fix an orientation of M with area form dM, and let 3D be a rec- 
tangular decomposition of M whose 2-segments x u • • • , x f are all positively 
oriented. Thus 2D is in particular an oriented paving of M as denned in 
Chapter VI, Section 7. By definition the total curvature of M is 



// KdM = ±J[ 



KdM 



(1) 



We shall apply the Gauss-Bonnet formula to each summand. (This is 
permissible, since on each region Xi(Ri) the area form dM is the one de- 
termined by x;. ) In terms of interior angles, this formula reads 

t A topological invariant is a property preserved by every homeomorphism (that 
is, continuous function with a continuous inverse). A diffeomorphism is a homeo- 
morphism, but not conversely. However it is a peculiarity of low dimensions that 
two surfaces are diffeomorphic if (and only if) they are homeomorphic. 



Sec. 8] 



THE GAUSS-BONNET THEOREM 



381 



// K dM = - / Kg ds -2x + ( 4l + i 2 + t 3 + l 4 ) 



(2) 



Now consider what happens when (2) is substituted in (1). 

Because M is a surface— locally like E 2 — each edge of the decomposition 
£> will occur in exactly two faces, say Xi(Ri) and xj(Rj). Let en and a, 
be the parametrizations of this edge occurring in dx» and dx h respectively. 

Because these regions have the same orientation as M itself, ai and « y 
are orientation-reversing reparametrizations of each other, as in Fig. 
7.45. Thus 



/ Kg dS + / Kg dS = 0. 



It follows that 



»=1 •'dx. 



Kg ds = 



(3) 



for we have just seen that the integrals over edge curves will cancel in 
pairs. (As usual, we write v, e, and / for the number of vertices, edges, 
and faces in the decomposition.) 

Thus the substitution of (2) in (1) yields 



I 



KdM = -2x/+ tf 



(4) 



where d is the sum of all interior angles of all the faces in the decompo- 
sition. But the sum of the interior angles at each vertex is just 2x (Fig. 
7.46), so d = 2irv. Thus 



I 



KdM = -2tt/ + 2ttv 



(5) 




FIG. 7.45 



382 RIEMANNIAN GEOMETRY [Chap. VII 




FIG. 7.46 



A simple combinatorial observation will complete the proof. The faces 
of the decomposition 3D are rectangular: Each face has four edges. But each 
edge belongs to exactly two faces. Thus 4/ counts e twice; that is, 4/ = 2e. 
Equivalently, — / = / — e, so (5) becomes 

ff KdM = 2x(y - e + /) = 2t X (M). | 

Because the Euler-Poincare' characteristic is a topological invariant, 
this theorem shows that total curvature is a topological invariant. 

Explicitly we let M and M be geometric surfaces that are merely diffeo- 
morphicf Then the Gaussian curvatures K and K of M and M can be 
quite different — but their total curvatures are identical, for (being diffeo- 
morphic) M and M have the same Euler-Poincare 1 characteristic; hence 



ff KdM = 2ir X (M) = 2ttx(M) = ff_K 



dM. 



We have already seen instances of this theorem. For example, the torus 
in Example 2.5 has K = 0, hence total curvature zero. Alternatively, this 
same surface acquires from E its usual geometric structure as a torus of 
revolution, for which the curvature is variable — but we found in Chapter 
VI, Section 7, that its total curvature is also zero. (The diffeomorphism in 
this case is just the identity mapping.) 

In general it suffices to count handles to find total curvature. 

8.9 Corollary If ¥ is a compact orientable surface with h handles 
(h = 0, 1, 2, • • • ), then for any geometric structure on M, the total curva- 
ture is 4x(l — h). 

Proof. We have already seen that M has Euler-Poincare" characteristic 
2 - 2h. I 

j See footnote, page 380. 



Sec. 8] THE GAUSS-BONNET THEOREM 383 

The Gauss-Bonnet theorem (Theorem 8.8 or 8.9) provides a way to attack 
some seemingly formidable problems. For example, (1) of Example 2.3 
shows that if one single point is removed from a sphere 2, there exists a 
geometric structure on the punctured sphere with K = 0. But there can be 
no geometric structure on an entire sphere 2 for which K ^ 0, since then 



//. 



KdL ^ 0, 



contradicting 2ttx(2) = 47r. Reversing this argument, we find that a 
compact orientable geometric surface with K > must be diffeomorphic to a 
sphere. Its total curvature is positive— but in Corollary 8.9, h is a nonnega- 
tive integer, so it must be zero. Thus the surface has no handles; it is diffeo- 
morphic to the sphere 2 = 2(0). Further results of this type are given in 
the exercises. 

The Gauss-Bonnet theorem is proved by cutting M into rectangular 
regions, and applying the Gauss-Bonnet formula to each. The scheme works 
because all these regions are consistently oriented by an orientation of M 
itself, so that the integrals Jk ds on the boundaries of these regions cancel 
in pairs. Here in essence is the fundamental idea of algebraic topology; 
indeed considerations of this kind led Poincare" to its invention (see Lef- 
schetz [8]). By applying this scheme to suitable regions in M we can get a 
more general form of the Gauss-Bonnet theorem (Exercise 8). A corollary 
(Exercise 11) shows how to extend Theorem 8.4 from rectangles to ar- 
bitrary polygons. To see how the notion of boundary generalizes in such situ- 
ations we shall give a direct (and logically unnecessary) proof of Exercise 
11 in the special case of a triangle, that is, the one-to-one regular image A 
of an ordinary triangle T in E 2 (Fig. 7.47). 

Corollary 8.10 If A is a triangle in a geometric surface M, then 
// KdM + f K g ds = 2tt — (e x + e 2 + e 3 ) = (n + n + n) - v. 
(We explain this notation in the course of the proof.) 




FIG. 7.47 



3*4 RIEMANNIAN GEOMETRY [Chap. VII 

y 




FIG. 7.48 



Proof. Let dM be an arbitrary area form on the region A. 

We get a rectangular decomposition of A = y(T) as follows. Cut T 
into three quadrilaterals, as indicated in Fig. 7.48; then changes of varia- 
bles in y will exhibit their images as rectangular regions xi(fli), x 2 (R 2 ), 
x 3 (R 3 ) constituting a rectangular decomposition of A. As usual we arrange 
for each x* to be positively oriented. Thus, by the Gauss-Bonnet formula 
with interior angles, the total curvature of A is 

ff K dM = £ ff K dM = - j^ f K.d8-br + g 

VJA »-l JJ %i i=l J dXi 

where d is the sum of all interior angles. 

Of the twelve edges in d Xl , dx 2 , and dx 3 , the six interior ones cancel in 
pairs (at least Jk s ds does, on them). The remaining six combine in pairs 
to give the curves a lt a 2 , <x 3 (Fig. 7.47) constituting the boundary dA of the 
oriented triangle A. Hence 



2 / K dS= Kg dS = Kg dS + / Kg dS + Kg 

i-l J dxi JdA J ai J a2 J az 



In the sum 0, the interior angles n, t 2 , t 3 at pi, ps, p 3 are those of the 
triangle A itself. The others, which occur at the artificially introduced 
vertices, evidently add up to 5tt; thus we find 

KdM + J K g dS = U + la + i,) - 7T. 
•'•'A JdA 

We are adapting to the triangle the definitions in 8.3; thus i, + e,- = ic 
gives the exterior angle formula. | 

If the edge curves of the triangle are geodesies, then of course the geo- 
desic curvature term vanishes. In particular, for a geodesic triangle in a 
surface with constant curvature K, this result reduces to 

ll + t2 + t3 = 7T + KA 

where A is the area of the triangle. Thus the well-known theorem of plane 
geometry that the sum of the interior angles of a triangle is ir depends on 



Sec. 8] 



THE GAUSS-BONNET THEOREM 



385 





the fact that E 2 is flat. Examples show easily enough how a geodesic tri- 
angle can manage to have n + i 2 + i 3 larger than jona sphere (K > 0) 
and smaller than x on a hyperbolic plane (K < 0) (Fig. 7.49). 



EXERCISES 

1. Find the total Gaussian curvature of: 

(a) An ellipsoid. 

(b) The surface in Fig. 4.10. 

(c) M : x 2 + y* + z« = 1. 

2. Prove that for a compact orientable geometric surface M : 
K > =* M is diffeomorphic to a sphere (text) 

K = => M is diffeomorphic to a torus 
i£ < ^ M is a sphere with /i ^ 2 handles 

3. (a) Let M be a compact orientable geometric surface with h handles. 

Prove that there exists a point p of M at which 

K(p) > if h = 0, 

K(p) =0 if h = 1, 

#(p) < if A ^ 2. 

(b) If M is a compact orientable surface in E 3 which is not diffeo- 
morphic to a sphere, show that there is a point p of M at which 
#(p) < (compare Theorem 3.5 of Chapter VI). 



386 RIEMANNIAN GEOMETRY [Chap. VII 

4. (a) For a regular curve segment a: [a, b] — » M, show that the 
total geodesic curvature /« k ds is 



/ 



dt. 



(Hint: Exercise 9 of Section 4.) 
(b) Let x be a (positively oriented) orthogonal patch in M. Deduce 
the following formulas for total geodesic curvatures of parameter 
curves : 

(Note that \E V = — x uu *x v and %G U = — x rc «x„, and if M is in E , 
either intrinsic or Euclidean derivatives give the same results.) 

5. Let x: R —> 2 be the geographical patch (Example 2.2 of Chapter IV) 
restricted to the rectangle R: ^ w, v ^ 7r/4. Compute separately each 
term in the Gauss-Bonnet formula for x. 

6. If F: M — * N is a mapping of compact oriented surfaces, the degree 
d F of F is the algebraic area of F(M) divided by the area of N. Thus 
d F represents the total algebraic number of times F wraps M around 
N. If M is a compact oriented surface in E 3 , prove the Hopf theorem: 
The degree d G of the Gauss mapping is the integer x(M)/2. (It can 
be shown that degree is always an integer.) 

An oriented polygonal region (P in a surface M is an oriented region that 
has a rectangular decomposition x x , • • • , x* which we always arrange to be 
positively oriented. Then the boundary d(P of (P is the formal sum of those 
edge curves appearing in exactly one of the boundaries dxi, • • • , dx k . We 
exclude the situation shown in Fig. 7.50, so d(P consists of simple closed 
(broken) curves. These definitions are such that if a is one of the edges 
in d(P then J (a) always points into the region (P. (This rigorizes the rough 
rule: "Travel around the boundary keeping the region always on the left.") 




FIG. 7.50 



Sec. 8] 



THE GAUSS-BONNET THEOREM 



387 




FIG. 7.51 



(a) If is a 1-form on an oriented polygonal region (P, prove the gen- 
eralized Stokes' theorem 



// d<t>== I 

J J<9 J d<5> 



0. 



(If d(P = 2 on, then f d(? <j> means S f ai <j>.) (Hint: Lemma 6.6 of 
Chapter IV produces some cancellation in pairs, as in the proof 
of Theorem 8.8.) 

(b) Deduce that if <f> is any 1-form on a compact oriented surface M, 
then jj M d4> = 0. 

(c) Two different (positively oriented) rectangular decompositions 
of the same region (P produce technically different boundaries of 
(P; however, both occupy the same set of points. Prove that for 
any 1-form on (P, f d6 > <f> is the same for both. 

(Generalized Gauss-Bonnet theorem). If (P is an oriented polygonal 
region in a geometric surface, prove that 

ft KdM + f K g ds + Y, e i = 2tx(<P) 

where 2 e> is the sum of the exterior angles of (P as defined in Definition 
8.3 for the special case of a rectangular region (Fig. 7.51). 

(Hint: Refine the proof of Theorem 8.8, classifying edges and vertices 
into those in d(P and those in the interior of (P. Note that on each simple 
closed boundary curve, the number of edges is the same as the number 
of vertices.) 

Prove that the following properties of a compact orientable surface M 
are equivalent: 



388 RIEMANNIAN GEOMETRY 



[Chap. VII 




FIG. 7.52 



(a) There is a nonvanishing tangent vector field on M 

(b) X (M) = 0. 

(c) M is diffeomorphic to a torus. 

(Hint: For (a) => (b), introduce a geometric structure and use Exer- 
cise 7.) Properties (a) and (b) are, in fact, equivalent for any compact 
manifold. 

10. (a) If a region (R has a rectangular decomposition, derive a triangular 

decomposition, and show that v - e + / is the same for both. 
(b) Do the same with "rectangular" and "triangular" reversed. 

The notion of simple region (Ex. 12 of VI.7) may be extended by allow- 
ing the mapping F: D -* M to be nondifferentiable (but still continuous) 
at n points on the circle u 2 + v 2 = 1. This permits n corners to appear in 
the boundary d(P of (P = F(D). We call (P in this case an n-polygon (n 
^ 0) The Euler-Poincare characteristic of an n-polygon is +1, since 
for a triangular decomposition as in Fig. 7.52, we have v - 1 = f = e /2 

11. If (Pis an oriented geodesic n-polygon (the edges are geodesies) in 
a geometric surface, show that 



IL 



K dM = 2t - £ e k = (2 - n ) T + £ t . 



y-i 



where e y and i,- are the exterior and interior angles of (P. 

12. (Continuation), (a) If (P is a geodesic n-polygon in the plane, prove 

that n ^ 3 and that the sum of the exterior angles of (P is 2x. 

(b) If M is a surface with constant Gaussian curvature K ^ 0, show 
that the area of a geodesic polygon is determined by its exterior 
or interior angles. 

(c) In the sphere 2 of radius r, find a geodesic 3-polygon (P whose 
interior angles are each 3tt/2. What is the area of (P? 

13. (a) In a surface M with K ^ prove that there exist no geodesic 

n-polygons with n ^ 2. Thus, in particular, two geodesies in M 
cannot meet to form the boundary of a simple polygonal region. 



Sec. 9] SUMMARY 389 




FIG. 7.53 

(b) On a sphere 2, for which values of n ^ do there exist geo- 
desic w-polygons? (Do not count "removable vertices" — those 
with exterior angle 0. ) 

14. In the hyperbolic plane, let (P B (n ^ 3) be a "geodesic n-polygon" 
whose vertices are on the rim u 2 -J- v = 4 of H, hence are not actually 
in H (Fig. 7.53). Find the area of (P n . 

15. (Hopf Umlaufsatz) . If /3 is a simple closed curve in E 2 , then the total 
geodesic curvature of is ±2x. Thus the unit tangent T of /3 turns 
through one full circle in traversing /?. Prove this result assuming 
/3 is the boundary curve of a simple region S. (Hint: S is a 0-polygon 
as defined in the paragraph preceding Ex. 11.) 

The assumption above is always true, but its proof requires rather deep 
topological methods. 



9 Summary 

A geometric surface — that is, a 2-dimensional Riemannian manifold — 
generalizes the Euclidean plane by replacing E 2 by any surface and re- 
placing the dot product on tangent vectors by arbitrary inner products. 
In the resulting Riemannian geometry, the length of a curve is defined as 
before, and gives a notion of intrinsic distance directly generalizing the 
familiar Euclidean distance in the plane. The acceleration of a curve is also 
a geometric notion, but it is not so obvious how an inner product on tangent 
vectors can lead to a measurement of the turning of a curve. For 70 or 80 
years after Riemann this was accomplished by rather complicated formulas 
in terms of coordinate patches (4.2 is a sample). In the Cartan approach, 
the inner products serve to define the notion of frame field, and the rate 
of rotation of a frame field is expressed by its connection form. The con- 
nection equation V y #i = o} n (V)E 2 then defines the covariant derivative— 
with acceleration as a special case. 

In Riemannian geometry as in Euclidean geometry, geodesies are the 



390 RIEMANNIAN GEOMETRY [Chap. VII 

curves with acceleration zero. Geodesies are not only straightest curves, 
however; they are also shortest curves in the sense discussed in Sections 
5 and 6. The simple Euclidean rule that "a straight line is the shortest 
distance between two points" is no preparation for the new and subtle 
behavior of geodesies on an arbitrary geometric surface — or even a surface 
as simple as a sphere or a cylinder. Some idea of how far the analysis of 
geodesies can lead may be found in Milnor [7]. 

It is by now hardly necessary to say that the Gaussian curvature K of a 
geometric surface M is its most important geometric property, for we have 
seen that sooner or later curvature enters into almost every geometrical 
investigation. Indeed K could be defined, for example, in terms of parallel 
vector fields (holonomy) or radial geodesies (the Jacobi equation) or 
polar circles. (For a surface in E 3 we used the shape operator, and could 
have used the Gauss mapping.) In the Cartan approach, however, curva- 
ture is defined by the structural equation duu. = —K6 X * 2 , which pre- 
sents K (in a sense discussed earlier) as the common "second derivative" 
of all frame fields on M . And it is this definition that leads most directly 
to the central result of two-dimensional Riemannian geometry, the Gauss- 
Bonnet theorem. Leaving aside trigonometric consequences such as Corol- 
lary 8.10, the content of the theorem is that curvature determines topol- 
ogy, at least in the compact orientable case. 

Generally speaking, the results of this chapter are valid for Riemannian 
manifolds of arbitrary dimension n, and in most instances the definitions 
and proofs need scarcely any readjustment. Dimension 2 has simplified 
certain consistency proofs such as Theorems 2.1 and 3.2, but these can 
be avoided entirely by more advanced methods. As one might expect, it is 
the Gauss-Bonnet theorem whose generalization offers the most difficulty 
(see Hicks [5]), and in higher (even) dimensions the curvature of M in- 
fluences but does not control the topological configuration of M. 



Bibliography 



1. H. Flanders, "Differential Forms: With Applications to the Physical Sciences." 
Academic Press, New York, 1963. 

2. G. Birkhoff and S. MacLane, "A Survey of Modern Algebra." Macmillan, New 
York, 1953. 

3. T. J. Willmore, "An Introduction to Differential Geometry." Oxford Univ. Press, 
London and New York, 1959. 

4. R. Courant and H. Robbins, "What is Mathematics?" Oxford Univ. Press, London 
and New York, 1941. 

5. N. J. Hicks, "Notes on Differential Geometry." Van Nostrand, Princeton, New 
Jersey, 1965. 

6. D. J. Struik, "Lectures on Classical Differential Geometry." Addison-Wesley, 
Reading, Massachusetts, 1961. 

7. J. W. Milnor, "Morse Theory." Princeton Univ. Press, Princeton, New Jersey, 
1963. 

8. S. Lefschetz, "Introduction to Topology." Princeton Univ. Press, Princeton, New 
Jersey, 1949. 

The books of Willmore and Struik are at about the same level of difficulty as this 
one. Hicks' book begins at a level comparable with our Chapter VII and gives a very 
concise exposition of multidimensional Riemannian geometry; its bibliography lists a 
number of more detailed works on the subject. 



Answers to Odd- 
Numbered Exercises 

(These answers are not complete; and in some cases 
where a proof is required, we give only a hint.) 



Chapter 



Section 1 



1. (a) x*y* sin 2 z, (c) 2x*y cos z 

3. (b) 2x<* cos (e*), h = x 2 + y 2 + z 2 

Section 2 

1. (a) -6C7i(p) + U»(p) - 9C7 3 (p) 
3. (a) V = (2z*/7)Ui - (xy/7)U 3 
(c) V = xUi + 2yU 2 + xy*U 3 
5. (b) Use Cramer's rule. 

Section 3 

1. (a) 0, (b) 7-2 7 , (c) 2e 2 

3. (a) y\ (c) yz*(y 2 z - Sx 2 ), (e) 2x(y* - 3z 5 ) 

5. Use Ex. 4. 

Section 4 

1, «'(x/4) = (-2,0,V2)p, where p = (1,1,V2) 
3. /3(s) = (2(1 - s 2 ), 2s Vl - s 2 , 2s) 
5. The lines meet at (11,7,3) 
7. v„ = (1,0,1), 
9.Ata(0); «-> (2,2*,*) 

Section 5 

1. (a) 4, (b) -4, (c) -2 

5. (b) (xdy - ydx)/(x 2 + 2/ 2 ) 

393 



394 ANSWERS TO EXERCISES 

7. (a) dx — dz, (b) not a 1-form, (c) zdx + x dy, (d) 2(xdx + ydy), (e) 0, (f) not 

a 1-form 
9. ± (0, 1, f) 
11. (a) Use the Taylor approximation of the function t — >/(p + t\) 

(b) Exact: —0.420, approximate: — \ 

Section 6 

1. (a) <f> A $ = yz cos z dx dy — sin z dx dz — cos z dy dz 

(b) d^ = —zdxdy — ydxdz. Note that d(dz) = d(l-dz) = Oby 1.6.3. 
7. Apply this definition to the formula following 1.6.3. 
9. Assuming the formula, set / = y, g — x. 

Section 7 

1. (a) (0,0), (b) (-3,1), (3,-1), (c) (0,0), (1,0) 
5. (a) (2,0,3) at (0,0,0), (b) (2,2,3) at (0,2,tt) 
7. GF = (flritfu/O.fcC/i,/.)) 
11. (a) F-* = (», ue~°), (b) F- 1 = (m 1 ' 3 , t> + « 1/3 ), (c) F" 1 = ((9 - u - 2»)/2, 5 - 

tt — i>). F is a diffeomorphism for (a) and (c) only, since in (b), F~ l is not differ- 

entiable (when u = 0). 



Chapter II 



Section 1 



1. (a) -4, (b) (6,-2,2), (c) (1,2,-1)/V6, (-1,0,3)/Vl0, (d) 2^11, (e) -2/V15 
5. If v X w = 0, then u»v X w = for aK u; use Ex. 4. 
7. V2 = v — (v»u)u 

Section 2 



3. 0(8) = (VI + s 2 /2, *A/2, sinh" 1 (*A/2)) 

5. If /Si is based at ti{i = 1,2), then so is plus or minus the arc length of a from t\ to 

fe. 
9. (b) The condition is certainly necessary; to see that it is sufficient, show that a 
unit speed reparametrization of a. has acceleration zero. 

11. (b) L(a) > / a'-u dt = Y, -r 1 Uidt = (q - p)»u = d(p,q) 
•>« J« * 

Section 3 

1. K = l, T = 0, B = (-|,0,-|), center (0,1,0), radius 1. 
7. (a) 1 = || a(h)' || = || a'(h)h' || = | h' |, hence A = ±1 

(b) Let e = ±1, then a = a(h) implies T = a' (h)h' = eT(h); hence 

«iV = k(/i) N(h), and so on. 
9. The orthogonal projection on the N B plane (the normal plane of /3 at /3(0)) is 

s -» Ko (sV2)iVo + k t (s 3 /Q)Bo (cusp at s = 0). 



ANSWERS TO EXERCISES 395 

1 1. B = B implies tN = fN, hence either (1) f = r and N = N, or (2) f = — r and 

N N. 

Section 4 

1. Let/ = t* + 2; then K = r = 2//»; 5 = {t\-2t,2)ff 

3. (a)JV(O) = (0,-1,0), r(0) = f 

9. (a) * = x/4, u = (1,0,1) A/2, t(0 = « - (*76), < 2 , -« + « 3 /6)) 
15. (c) The evolute is also a cycloid. 
17. «'(*) = (/(*) sin «,/(*) cos t,f(t)g(t)) 

Section 5 

1. (a) 2tfi(p) - t/ 2 (p), (b) C/^p) +2[/ 2 (p)+4 C7 3 ( P ) 
5. (a) 8tf,(p) - 4(t/ 3 )(p) 

Section 6 

1. Show that V'W = 0, and use III .8. 

3. For instance, E 2 = —sin 2t/ 2 + cos zU 3 , and E z = E t X E 2 . 

Section 7 

1. W12 = 0, W13 = W 2 3 = (d/)/\/2 

3. wi2 = — df, con = cos / df, w 2 3 = sin f df 

5. By (3) of II.5.4, V v (E/^i) = £ Ff/;]^ + E/iV^-. 

7. At an arbitrary point p, a(t) = fp is a curve with «' = [| p [| Fi. Show that 

II P II Fi\p] = II P ||. 

Chapter III 

Section 1 

3. (T a )~ l = T_ a , CM = -«C, hence F" 1 = (7\,C)-i = C- 1 ^ = TVc-i^C" 1 . 
5. (b) Using Ex. 3, we find F~i(p) = (5y/2, -5, 4V2) 

Section 3 

1 . If F and G have orthogonal parts A and B, then by III. 1 .2 , sgn (F(?) = det (AB) 

= det ^-det B. 
5. C is a rotation, through angle ir/2, around the axis given by a. 

7. For E 1 : F(s) = es + s ; for E 2 : F = 



m /cos # -« sin #\ 

:F=TC, where C = ( J. 

\sin # e cos #/ 



Section 4 



1. (b) By definition, p(s) is the point canonically corresponding to T(s), hence by 

III.2.1, 0(0) corresponds to ^(T 7 ), the unit tangent of F(0). 
5. For a tangent vector v at p, F* (V V W) = W(F(p) + *C(v))'(0) = V ,+ M W. 



396 ANSWERS TO EXERCISES 

Section 5 

1. P = T p (C(a)), where C( Ui ) = e, 

3. Consequence of III.5.7 

5. If t is not identically zero, assume r (0) ^ and examine the proof of 5.3. 

7. Let i? = TC, where T is translation by (0,0,bs /c) and 

(cos (s /c) -e sin (s /c) 0\ 
sin (s /c) e cos (s /c) J 
J 

where e = ±1. Then F(/8) = /3(«s + s„) 
9. a(s) = (/ cos <p(s) ds, / sin *>(s) ds), where *>(«) = / k(s) ds 
11. Use Ex. 9 



Chapter IV 
Section 1 

1. (a) The vertex, 0, (b) all points on the circle x* + y* = 1, (c) all points on the z- 

axis 
5. (b) c ^ -1 
9. Use Ex. 7 
1 1. q is in F(M) if and only if F~ l (q) is in M, that is, g(F~Hq)) = c. Use the hint 

to apply IV.1.4. 

Section 2 

1. (c) x(u,v) = (u,v,v? + v 2 ) is one possibility; a parametrization derived from 

IV.2.5 will omit a point of the surface. 
5. x u X x v = vd' X 5 

7. (b) Straight lines (rulings) and helices, (c) M: x sin (z/b) = y cos (2/6) 
9. x(w,y) = (cos u — v sin w, sin w + v cos w, i>) 
13. (a) If g' is never 0, reparametrize the profile curve to obtain u — >■ (u,f(u),0), 

and use IV.2.5. 

Section 3 

1. (a) r 2 cos 2 v, (b) r 2 (l — 2 cos 2 v cos w sin u) 

3. (a) m and v are the Euclidean coordinate functions of x x y 

(b) Express y = x(u,v) in terms of Euclidean coordinates, and differentiate. 
5. (a) M is given by g = z - /(ar,y) = 0, with Vj = (-/„-/ y ,l), and v is tangent 

to M at p if and only if vV^(p) = 0. 
7. Vg = (— 2/,— a;,l) is a normal vector field; V is a tangent vector field if and only 

if V*Vg = 0, for example, V = (x,0,z). 
9. (a) T P (M) consists of all points r such that (r — p). z = 0; hence v p is in T P (M) 

(that is, vz = 0) if and only if p + v is in T P (M). 
11. (a) 2tt 



ANSWERS TO EXERCISES 397 

Section 4 

= (d/ a + /dtf>) (x„ , xj 
5. If a is a curve with initial velocity v at p, then 

yp[g(f)] = (ftf«)'(0) = </'(/«) (0)(/«)'(0) = 0'(/(p) )▼*[/]• 

7. On the overlap of ^f and 11 ,, d/» - df, - d(/< - //) = 

9. (b) d«t(x„) = x„M = — — = 1. 

du du 

Section 5 

1. If x: D -» M is a patch, then F(x) : D -► JV is (by 3.2) a differentiable mapping. 

Hence y -1 Fx is differentiable for any patch y in AT. 
3. If x and y are patches in M and N, respectively, then y _1 Fx = (y -1 y) (x -1 x) is 

differentiable, being a composition of differentiable functions. 
7. (b) x*(»») = r 3 sin 2 v cos v du dv 
11. Only (a) is not a diffeomorphism. 
13. (b) F*(ax u + bx v ) = ay„ + by v implies linearity. 

Section 6 

7. (a) 2irw, (b) 2-jrn 

13. (b) Show f a <\> = Jx d<t> for a suitable x. 
15. Use simple connectedness to prove the formula given in the hint — see Fig. 4.46. 

Section 7 

1. (a) Connected, not compact, (c) connected and compact, (e) connected, not 

compact. 
3. If v is nonvanishing on N, show that F*v is nonvanishing on M . 
5. (a) If Z is a nonvanishing normal, then let ±U = ±Z/ || Z \\. If V is an arbitrary 

unit normal, write V = (V'U)U and use Ex. 4(b). 

(b) The image x(D) of a patch with D connected. 
9. (c) Use Ex. 7 

Section 8 

1. Modify the proof given for the Mobius band in IV.7. 

9. (x X y) _1 (x X y) = (x _1 x) X (y _1 y), a differentiable function. 



398 ANSWERS TO EXERCISES 

Chapter V 

Section 1 

1 . Use Method 1 in text. 
3. (a) 2, (c) 1 

Section 2 

1. (b) If ei, e 2 = (ui ± u 2 )/V2, then £( ei ) = ei and S(e 2 ) = -e 2 

Section 3 

5. (b) An ellipse on one side and no points on the other; the two branches of a 
hyperbola (asymptotes the two lines in (a)); two parallel lines on one side, and 
no points on the other. 

7. (a) If a is a curve in M with initial velocity v at p, then F*(v) = F(a)'(0) = 
(a + eU a )'(0) = v - eS(v) at F(p). 

Section 4 

5. K = -36r 2 /(l + 9r 4 )*; not minimal 

7. Compute speed from a' = a/x M + a 2 'x v 
13. p = x(u,v) is umbilic if and only if S(x u ) = Ax„ and S(x v ) = Ax, at (u,v). Dot 

with x„ and x v . 
15. (a) None, since K < 0, (b) origin (a planar point), 



/ b a 2 _ ^ 2 \ 

(c) I 0, ± - Va 2 - 6 2 , — - — ) if a > b 



Section 5 

3. A meridian a lies in a plane orthogonal to M along «, hence a" is tangent to this 
plane, and (with constant speed parametrization) orthogonal to «'; thus a" is 
orthogonal to M. 
7. S(T) = —U', hence by orthonormal expansion, U' = — S(T)»T T — S(T)»V V. 
Continue as in the proof of the Frenet formulas. 

15. On the ruling through <r(u), the formula for K in Ex. 14 shows that either K is 
identically zero, or K < has minimum value —\/p(u) 2 at a(u) and rises sym- 
metrically toward zero as v — * ± °° 

17. (a) For (w,0,0) + v(0,l,u): the x axis, with p(u) = 1 + u 2 . (b) Use Ex. 15. For 
fixed u, K = - (1 + m 2 + ^)-2 is a minimum when v = 0. 

19. (c) x = a + v8 is noncylindrical, and we can assume that a is a (unit speed) 
striction curve. But a'»S X 5' = (since K = 0) and a'-S' = (striction), hence 
T = a! and 5 are collinear. 

Section 6 

T. K = (1 - z'Xl + x* exp (-a; 2 ))- 2 

3. Use the results of V.2. Note that meridians are normal sections. 

5. M has parametrization x(u,v) = (u cos v, u sin v, /(«)). 



ANSWERS TO EXERCISES 399 

7. With the usual parametrization, an argument like that for V.6.2 reduces to the 
extreme cases: g' always zero, g' never zero. In the former, M is part of a plane 
(special case of cone). 

9. (c) If y = f(x) has unit speed parametrization (g,h), where h(u) = ce~ ulc , show 
that / (not h !) satisfies the differential equation in VI .6.6. 



Chapter VI 

Section 1 

1 . (a) a" = ui2 (T)E 2 + a>i3(T)E 3 , hence a" is normal to M if and only if ui 2 (T) = 0. 
5. If the cylindrical frame field is restricted to M and the indices 1 and 3 are reversed, 
we obtain the frame field in (1) of VI.1.3. By the computations in II.7, 

0>12 = «13 = 0, 0>23 = — d& 

Section 2 

3. (a) yp = *CEi)«i + ^(E 2 )e 2 = -Mi + fad 2 . 
(b) f = hio) 23 — hvmz 

Section 3 

I . If K = H = 0, then fafa = h + k 2 = 0. Thus h = fa = 0, and S = 0. 

3. Assume fa 9^ fa, and use the Hilbert Lemma (3.6) to get a contradiction. 

5. In the case fa ^ fa, use VI.2.7 to show that, say, fa = 0. By Ex. 2 the k\ princi- 
pal curves are straight lines. Show that the fa principal curves are circles, and that 
the (&i) straight lines are parallel in E 3 . 

Section 4 

1. (d) => (b): if u is an arbitrary tangent vector at p, then u = av + b-w, hence 

|| F*u || 2 = a 2 || F*v || 2 + 2ab F*vF*w + 6 2 || F*w || 2 = 

a 2 || v || 2 + 2ab vw + fc 2 \\ w || 2 = || u || 2 

3. If a is a curve segment from (—1,0,0) to (1,0,0) with length 2, then by the exercise 
mentioned, a parametrizes a straight line segment — impossible, since a must 
remain in M. 

5. (a) Define F( a (u) + vT a (u)) = 0(u) + vT„{u) 

(b) Choose fi in E 2 with the same curvature function. 

7. (a) Criterion (a) becomes F*{\)*F*{\f) = X 2 (p)v»w; criterion (c) becomes 

n( ei ).F*( e/ ) = x 2 ( P )5 iy . 

II. Write F(\(u,v)) = x(a(u), b(v)) for suitable parametrizations. 

13. For y, show that the conditions E = G and F = are equivalent to g' = cos g, 
which has solution g(v) = 2 tan -1 (e v ) — (ir/2) such that g(0) = 0. Use Ex. 7. 

15. F(x(u,v)) = (f(u) cos v,f(u) sin v), where x is a canonical parametrization and 
f(u) = exp ( /? (dt/h(t)). 



400 ANSWERS TO EXERCISES 

Section 5 

1. If a' = Ex along a , then F( a )' = F* (Ex) = Ex along F( a ). Use VI.5.3. 

3. There is no local isometry of the saddle surface M (-1 < K < 0) onto a catenoid 
with — 1 < K < 0, since K has an isolated minimum point (at 0) while K takes 
on each of its values on entire circles. (Many other examples are possible.) 

5. (b) Follows from VI.4.5, since for x t we compute E t = cosh 2 u = G h and F t = 
(independent of t). 

(d) For M t :U t = (s,-c,S)/C, so the Euclidean coordinates of U t are independent 
of t. 

Section 6 

1. (b) Ox = Vl + u 2 du, 02 = u dv, «i2 = dv/Vl + u 2 , K = 1/(1 + u 2 ) 2 . 

3. «12 = — t? u rft< 

Section 7 

3. (a) yl = (2tt/3){(1 + c 2 ) 3 ' 2 - 1}, (b) <*> 

5. (a) Use a canonical parametrization; then x*(KdM) = (-h" /h)(h du dv) = 

— h" du dv. Recall that h' = sin <p. 

(c) for the bugle surface, lim <p a = —1, lim <p b = 0. 

a-»0 6-*oo 

7. Use Ex. 5. At the rims of all these surfaces, h' — sin*?— ► ±1. For K = 1/c 2 (V.6.5): 
in Case (2), TC = <Lwa/c, A = 4irac; in Case (3), TC = 4tt, A = 4ttc 2 . For if = 
-1/c 1 (Ex. 9 of V.6) : M a has TC = 2x(a - c)/c, M b has TC = -4,r. 

9. (a) Use Ex. 14 of V.5. 



L 



pi 

dv = 2. 



( p 2 + ^)3/2 



11. Define x(u,v) = F((l - u) cos i;, (1 - u) sin »), on #: < u < 1, < » < 2tt. 

13. With outward normal, H = — 1/r, and h = r. 

1 5. Dividing if necessary by the lengths of V and W, we can assume they are unit 
vector fields. Thus V, U X V is a tangent frame field on a. Orthonormal expan- 
sion shows that if / = W*V and g = W*U X V, then/ 2 -f- g 2 = 1. 

17. (a) Use || ^v X F*w || = | J | || v X w || 

19. (a) F carries positively oriented pavings of M topositively oriented pavings of N. 
(b) If F: M — ► M is an isometry, orient M and M so that F preserves orientation 
(see remark following Ex. 4). Then 

JJ KdM = ff F*(K dM) = ff K(F)F*(dM) = ff K dM 

Section8 

1 . If A^ is isometric to a sphere 2 of radius r, show that N is compact and has K = 

1/r 2 . Thus,vby Liebmann's theorem, N is also a sphere of radius r, so a translation 

will show that N is congruent to S. 
3. Except for geodesies, all follow immediately from preservation of shape operators. 

Only geodesies and Gaussian curvature need be preserved by arbitrary isometries 

(Ex. 1 of VI.5, and theorema egregium). 



ANSWERS TO EXERCISES 401 

5. (b) That given by - U. 

9. Deduce from theorema egregium that a Euclidean symmetry F of M must leave 

the origin fixed, hence F is orthogonal. Consider its effect on the natural frame 

at 0. 



Chapter VII 



Section 1 



1. (b) <x'°a' = a'*a'/g 2 (a) is speed squared. 

3. (b) Let Ei and Ez be the vector fields on x(D) determined by x u /y/E and V / || V \\, 

where V = x v — (F/E)x u (Gram-Schmidt process). 
5. (a) Show that the definition J(Ei) = E 2 ,J(E 2 ) = -Ei is independent of the choice 

of positively oriented frame field. (Any two have the same orientation in the sense 

defined just preceding VII.1.4.) 

(b) For/ 2 = -/, show J(J(Ei)) E t . 

7. (a) F*£7, = f u Ui + gJJ* , F*t/ 2 = f t Ui + g v U, (Use Ex. 6). 

(b) The complex derivative of F = / + ig is dF/dz = /„ + ig« ■ Thus the magnitude 

| dF/dz | is (f u 2 + g u 2 ) 112 , which may be rewritten in various ways using the Cau- 

chy-Riemann equations. 



Section 2 

1. 0i = du/v, &2 = dv/v 

3. A = 7rr 2 /f 1 - — J , (E, F, G computed as in Example 4.11); A(H) = <». 

5. E = G = 1 and F = 0, so A = 4ir 2 . We can redefine the geometric structure to make 

E and G any positive numbers. 
7. Show that the parametrization x in Ex. 5 of VI.5 is an isometry. 

Section 3 

1. Note that E, = vUi restricted to a is r sint [/,- , thus a' = —E x + cot (0^2 . Since 
wis = du/v evaluated on a' is — 1, we derive from the covariant derivative for- 
mula that a" = cot (t)Ei - cot 2 (t)E 2 . 

3. In the proof referred to, note that u> l3 (V)E 3 = -V v EfEiE 3 = S(V)'EiE 3 . 

5. (a) The proof of VII.3.6 shows that the holonomy angle of a is 

«6 



— / Wl2(a') dt = — I C012 



Use Stokes' theorem, recalling that dwvi = —K dM (first structural equation). 
7. Y' = /'_Ei + fw2i{oc')E 2 , hence F„(F') = f'E, + }un{a')Et . F.Y = Y = /£, , 

hence F' = /'F, +/»ii(F,a')^i • Use VI.5.3. 
11. H 7 has constant length c; on any orientable region, show that the frame field 

Ei = H'/c, F 2 = /(Fi) has connection form zero, hence curvature zero. 



402 ANSWERS TO EXERCISES 



Section 4 



1. Since a" = 0, a{h)" = a '(h)h" 

3. Apply Ex. 6 of Section 3. 

5. All Euclidean circles through the north pole, since these correspond under sten- 
ographic projection to straight lines in the plane. 

9. Using the equations for a.' and a." given in the text, we have J (a') = J (vT) = vN 
hence a" •/(«') = Kg v 3 . 

11. (a) For u <u < Ul - t> a/ = ^° _ & > A t > 

y/EG ~ 

(b) // the meeting takes place, then <*' and 0' are collinear, hence by VII.4.2, « 

and are equal (but for parametrization)— an impossibility. 
13. (a) for the usual parametrization of a surface of revolution, G = h 2 (h distance 

to t'he axis of revolution), and the w-parameter curves are the meridians, hence 

the slant c is h sin <p. 

(b) Follows from Exs. 11 and 12, since such a parallel is a barrier curve. 
15. Meridians obviously approach the rim (in one direction). Use Ex. 13 to show 

that any other geodesic meets one barrier curve and approaches the rim in both 

directions. 
17. | sin <p j < 1/cosh m . 
19. (a) Use VI.2.4, (b) Kl = 0, k 2 = h'/h. 



Section 5 

1. p(0, p) = tanh _1 (|| p ||/2), Euclidean norm. 

3. The geodesies are helices, and y(u, v) = y U e 1+ ve 2 (l) = (r cos u/r, r sin u/r, v). 
The largest normal neighborhood is r, rr ; y is regular for any «, but the one-to- 
one condition fails for e > irr. 

5. (a) If q is the r, e , then by 5.5, p(p, q) < e . If q is not in r, t a curve from p to q 
meets Cs for every S < e. 

(b) If p ^ q, it follows from the Hausdorff axiom that there is a normal neighbor- 
hood of p which does not contain q, hence p(p, q) > 0. 

7. The length L of a from to r + e is the same as that of the broken curve: from 
to r, then a. from r to r + *. Hence L > p(a(0), a (r + e)) by the remark preced- 
ing 5.8. 

9. D: u 2 + v 2 < 1 with Euclidean geometric structure. 
11. (a) There is only one geodesic segment (a meridian) from p to any other point. 



Section 6 

5. (a) x u (0, v) = X(v), and since x(0, v) = 7xoo(0) = p(v), we have x„(0, v) = 0'(v). 

Thus EG - F 2 is nonzero when u = 0, and by continuity for | u \ small. 

(b) (iii) a base curve, X = S. 
7. (a) The y-parameter curves are meridians of longitude. 

(b) Since K = 0, the Jacobi equation becomes (VG)«« = 0, hence y/G is linear 

in u, and it follows that G(u, v) = 1 — Kg (v)u. 



ANSWERS TO EXERCISES 403 

Section 7 

1. If N is geodesically complete, fix a point p of M ; then there is a geodesic segment 

y w from F(p) to an arbitrary point q of AT. But for v such that F*(v) = w, we 

have F(y v ) = y w , hence q is in F(M). 
3. If p 5* q in M, then any geodesic segment a from p to q has nonzero speed. Hence 

F(<r) is a nonconstant geodesic from F(p) to F(q), and it follows that F(p) ^ 

F(q), even if "two" is interpreted as "two distinct" in the uniqueness property 

of TV.' 
5. Variant of VII.7.3. 
7. To show that a surface M is not frame homogeneous, it suffices (by the argument 

given in the text for the cylinder) to show that M has some geodesies which are 

one-to-one and some which are not. Note that x in VII.2.5 is a local isometry. 
9. If L is the arc length of the ellipse, show that F(x(u, v)) = x(u + (L/4), i>) is 

an isometry of M which does not preserve Euclidean distance. 
11. By Ex. 8 (b), F is the restriction to 2 of a linear (orthogonal) transformation, 

hence F(-p) = -F(p). Thus we_can define F{p, -p) = {F(p), F(-p)}, and 

deduce the frame-homogeneity of 2 from that of 2. 
13. The function is a homomorphism, so it suffices to prove it is one-to-one. But in 

the proof of VI.8.3, if M is not planar (so S ^ 0), then F is unique. 

Section 8 

1 . In (a) and (c) the surface is diffeomorphic to a sphere, so TC = 4*-. In (b) there 

are four handles, so TC = —12w. 
3. (a) For h = 1 : if K were never zero, then by an earlier exercise, either K > or 

K < 0; both are impossible, since by VII.8.9, //m K dM = 0. 

(b) By VI.3.5, K is somewhere positive. But M has at least one handle, so 



//. 



K dM < 0; 

M 



hence K is somewhere negative. 



5. J"/* KdM = - $to.K ds = x/4\/2 

7. (a) If xi , • • • , Xjt is a positively oriented decomposition, then by Stokes theorem 
(IV.6.5) we have /<p d<t> = 2 JUi d<t> = 2 /<*,- 4>- But for edges not in d(P, IV .6.6 pro- 
duces cancellation by pairs. 

(c) If Xi and y; are two positively oriented decompositions, then by the remark 
following VI.7.5, we have 

Jdyj JJ yj JJ *i J 3*i 

9- ( a ) => (b) : Any geometric structure onM (such exist) can be modified so that the 

nonvanishing vector field Fi has unit length; then Fi , J(E X ) is a frame field. 

Its connection form is defined on the entire surface, so 2ttx(M) = JJm K dM = 

— Jf M do>i2 = 0. 
1 1 . Consequence of Ex. 8. 
13. By the uniqueness of geodesies, interior angles of a geodesic polygon cannot be 

±7r (that is, there are no cusps), so — ■* < e < ir. Use Ex. 11. 

(b) all n 5^ 1 (for n = 0, a great circle). 



INDEX 



Acceleration, 54, 68 

of a curve in a surface, 196, 324(2&r. 3) 

intrinsic, 320 
Adapted frame field, 246-251 
Algebraic area, 289-290, 386 (Ex. 6) 
All-umbilic surface, 257-259 
Alternation rule, 27, 47, 153 
Angle, 44 

coordinate, 210 

exterior, 374 

interior, 374 

oriented, 291 
Angle function, 50(Ex. 12), 295(-£c. 15) 
Antipodal mapping, 165 (Ex. 5) 
Antipodal points, 182 
Arc length, 51-52, 220(Ex. 7) 

function, 51 
Area, 280-286 

Area form, 283-284, 292, 309(#x. 4) 
Area-preserving mapping, 295(Ex. 16) 
Associated frame field, 276-277 
Asymptotic curves, 226-227, 230 
Asymptotic directions, 225-227 
Attitude matrix, 46, 88 

B 

Basis formulas, 251 
Bending, 268, 275(Ex. 5) 
Binormal, 57, 66, 69 
Boundary 

of a region, 386 

of a 2-segment, 170 
Bracket operation, 81 (Ex. 7), 195(Ex. 9) 
Bugle surface, 241-242, 244(Ex. 9), 282- 
283, 291 



Canonical isomorphism, 7-8, 44, 59 
Canonical parametrization, 238 
Cartan, E., 42, 92, 96, 303 
Cartesian plane, 306 



Cartesian product, 187 (Ex. 9) 
Catenoid, 236 

Gaussian curvature, 236, 239 
total, 287-288 

Gauss mapping, 296(2&c. 20) 

local isometry onto, 267-268 

as minimal surface, 236-238 

principal curvatures, 236 
Center of curvature, M(Ex. 6) 
Circle, 59, 62 

Clairut parametrization, 330-339 
Closed surface in E 3 , 181 (Ex. 10), 263n 
Codazzi equations, 249, 255 
Compact surface, 176-177, 259-260 
Complete surface, see Geodesically com- 
plete surface 
Composite function, 1 
Cone, 140(Ex. 5), 2Sl(Ex. 11) 
Conformal geometric structure, 305-306, 

SW(Ex. 1), 312-313 
Conformal mapping, 268-271, 310 
Conformal patch, 270(Ex. 7), 279(Ex. 2) 
Congruence of curves, 116-123 

determined by curvature and torsion, 
117 
Congruence of surfaces, 189, 297-303 
Conjugate point, 353-359 
Connected surface, 176 
Connection equations 

on Euclidean space, 86, 248 

on a surface, 248, 318 
Connection forms 

on Euclidean space, 85-90 

on a surface, 248, 272, 277, 306 
Conoid, 233 (Ex. 20) 
Consistent formula for a mapping, 165 

(Ex. 10) 
Coordinate angle, 210, 22Q(Ex. 8) 
Coordinate expression, 143, lQ5(Ex. 6) 
Coordinate patch, see Patch 
Coordinate system, l58(Ex. 9), 276-277 
Covariant derivative 

Euclidean, 77-80, 116CEc. 5), 189-190, 
321 



405 



406 



INDEX 



intrinsic, 318-326 

on a patch, 212, 324-326 
Co variant derivative formula, 91 (Ex. 5), 

189-190, 321 
Cross product, 47-49, 107, 110 
Cross-sectional curve, 138 
Curvature, see also Gaussian curvature 

of a curve in E 2 , 65(Ex. 8), 122-123 

of a curve in E 3 , 57, 66, 69 
Curve, 15 

closed, 169 

coordinate expression for, 144 

one-one, 20 

periodic, 20 

plane, 61 

regular, 20 

simple closed, 151ra 

in a surface, 144-145 

unparametrized, 20-21 
Curve segment, 5Q(Ex. 10), 167 
Cylinder, 129, 231 (Ex. 11) 

geodesies, 275(Ex. 2), S52(Ex. 8) 

parametrization, 137, 141 (Ex. 6) 
Cylindrical frame field, 82-83 

connection forms, 89-90 

dual 1-forms, 9Q(Ex. 4) 
Cylindrical helix, 72-75 

D 

Darboux, 81 
Degree 

of a form, 27, 152 

of a mapping, 386 (Ex. 6) 
Derivative of Euclidean vector field, 113 
Derivative map, 35-40 

of an isometry, 104 

of a mapping of surfaces, 160-161, 166 

of a patch, 149 (Ex. 4) 
Diffeomorphism, 38, 40(Ex. 11), 161 
Differentiability, 4, 10, 33, 143, 145-146 
Differential, 23-26 
Differential form 

closed or exact, 157 (Ex. 2), 173-175 

on E 2 , 156 

on E 3 , 22-31 

on a surface, 152-157 

pullback, 163 
Direction, 196 
Directional derivative, 11-15, 149 

computation of, 12, 25 



Director curve, 140 
Distribution parameter, 2Z2(Ex. 14) 
Domain, 1 

Dot product, 42-44, 46, 82, 210 
preserved by isometries, 105 
Dual 1-forms, 91-92, 248, 306 
Dupin curves, 208(Ex. 5) 

E 

E, F, G, \4Q(Ex. 2), 210, 3H(Ex. 8) 
Edge (curve), 170, 373, 377 
Ellipsoid, l42(Ex. 10) 

Euclidean symmetries, 303 (Ex. 10) 

Gaussian curvature, 217-219, 222(Ex 
19) 

isometry group, 365 

umbilics, 223 (Ex. 23) 
Elliptic hyperboloid, 142(Ex. 10), 221-222 
Elliptic paraboloid, 142(Ex. 11), 220(Ex. 

6) 
Enneper's surface, 221 (Ex. 12) 
Euclid, 335-336 
Euclidean coordinate functions, 9, 15, 24, 

33, 53 
Euclidean distance, 43, 49 (Ex. 2) 
Euclidean geometry, 112, 304-305, 308 
Euclidean plane, 5, 305, 335-336 

local isometries, 363-364 
Euclidean space, 3, 5 
Euclidean symmetry, 302-303, 365 
Euclidean vector field, 147 
Euler-Poincare" characteristic, 378-380 
Euler's formula, 201 
Evolute, 75(Ex. 15) 
Exterior angle, 374 
Exterior derivative, 28-31, 31-32, 154-155 



Faces, 377 

Flat surface, 207, 2Sl(Ex. 11), 263(#r. 2) 

Flat torus, 316, S17(Ex. 5), S70(Ex. 7) 

imbedding in E 4 , 369 
Focal point, 362 
Form, see Differential form 
Frame, 44 
Frame field 

adapted, 246 

on a curve, 120 

on E 3 , 82 

natural, 9 



INDEX 



407 



principal, 254 

on surface, 251, 306 

transferred, 272-273 
Frame-homogeneous surface, 366 
Frenet, 81 

Frenet apparatus, 63 (Ex. 1), 71 
Frenet approximation, 60-61, 65 (Ex. 9) 
Frenet formulas, 58, 67 
Frenet frame field, 57 
Function, 1-2 

one-to-one, 2 

onto, 2 



Gauss, 245, 274, 303 

Gauss-Bonnet formula, 375-377, 383-385, 

388(Ex. 11) 
Gauss-Bonnet theorem, 380-383, 387 (Ex. 

8) 
Gauss equation, 249 

Gaussian curvature, 203-207, 310-312, see 
also individual surfaces 
differentiability, 206 
formulas 
explicit, 206, 212, 217, 253, 278 
implicit, 205, 252 
Gauss mapping and, 289 
geodesic curvature and, 339(Ex. 19) 
geodesies and, 355-358 
holonomy and, 325 (Ex. 5) 
interval, 275(Ex. 3) 
isometric in variance, 273-275 
in Jacobi equation, 355 
polar circles and, 359-360 
polar discs and, 361 (Ex. 2) 
principal curvatures and, 203 
shape operator and, 203 
sign, 204-205 
Gauss mapping, 194-195, 289-291 
Geodesically complete surface, 263n, 
328-329 
shortest geodesic segments in, 348 
Geodesic curvature, 230(Ex. 7), 329-330, 
337 (Ex. 9) 
total, 372-373 
Geodesic polar mapping, 340-341 
Geodesic polar parametrization, 341-344 
Geodesies, 228-232, 326-363, see also in- 
dividual surfaces 
closed, 232(Ex. 13) 



coordinate formulas, 327, 333 

existence and uniqueness, 328 

frame fields and, 250(Ex. 1) 

length-minimizing properties, 339-359 

preserved by (local) isometries, 275 
(Ex. 1), 326, 362-363 

spreading of, 353-354 

on surfaces of nonpositive curvature, 
358, 388(Ex. 13) 
Geographical patch, 134^135 
Geometric surface, 305, 308 
Gradient, 32(Ex. 8), 50(Ex. 11) 

as normal vector field, 148, 216 
Group, 103 

Euclidean, 103 

Euclidean symmetry, 302(Ex. 7) 

isometry, 365 

orthogonal, 104 

H 

Halmos symbol, 9 

Handle, 379, 382 

Hausdorff axiom, 186(Ex. 5), 345n 

Helicoid, 141 (Ex. 7) 

Gauss mapping, 296 (Ex. 20) 
local isometries, 267, 275-276 
patch computations for, 213-214 
as ruled minimal surface, 227, 233 (Ex. 
22) 

Helix, 15-16, 58-59, 119 

Hilbert's lemma, 261 

Hilbert's theorem, 263 

Holonomy, 323-325 
angle, 323 
Gaussian curvature and, 325 (Ex. 5) 

Homogeneous surface, 366-368, 370 

Homotopic to constant, 175 

Hopf, 386, 389 

Hyperbolic paraboloid, 143 (Ex. 12), 220 
(Ex. 6) 

Hyperbolic plane, 315-316 

geodesic completeness, 351 (Ex. 1) 

geodesies, 334-336 

isometries, 371 (Ex. 14) 

local isometries, 364 

of pseudo-radius r, 317 (Ex. 4) 



Identity mapping, 99 
Image, 1 



40(1 



INDEX 



Image curve, 33 

Immersed surface, 187 (Ex. 10), 219, 368 
Improper integral, 285-286 
Induced inner product, 308, 313 
Initial velocity, 21 (Ex. 6) 
Inner product, 304, 316-317 
Integral curve, 186-187 
Integral of function, 286, 292 (Ex. 4) 
Integration of differential forms, 167- 
176, 283-297 

1 -forms over 1-segments, 167-169, 172- 
173 

1 -forms over oriented regions, 285, 
292 (Ex. 4) 

2-forms over 2-segments, 169-172, 
174 (Ex. 8), 297 (Ex. 21) 
Interior angle, 374 
Intrinsic distance, 264-265, 269(Ex. 3), 

351 
Intrinsic geometry, 271, 304 
Inverse function, 2 
Inverse function theorem, 39, 161-162 
Isometric imbedding, 367-369 
Isometric immersion, 367-369 
Isometric invariant, 271, 304 
Isometric surfaces, 265 
Isometry of Euclidean space, 98-111 

decomposition theorem, 101 

derivative map, 104 

determined by frames, 105 
Isometry group, 365 

Isometry of surfaces, 263-266, 270-275, 
306 

Euclidean isometries and, 297-299 

isometric immersions and, 367 
Isothermal coordinates, 279 (Ex. 2) 



J (rotation operator), 309(Ex. 5) 
Jacobian, 288, 294(Ex. 8), 296(Ex. 17) 
Jacobian matrix, 37 
Jacobi equation, 355-357, 361 (Ex. 6) 

K 

Kronecker delta, 23, 45 

L 

2, m, n, 211-213 

Lagrange identity, 20® (Ex. 6) 
Law of cosines, 370 (Ex. 10) 



Leibnizian property, 13 
Length 

of a curve segment, see Arc length 

of a vector, 44 
Levi Civita, 322 
Liebmann's theorem, 262 
Line of curvature, see Principal curve 
Liouville parametrization, 339 (Ex. 18) 
Local isometry, 265-270, 362-365 

of constant curvature surfaces, 364 

criteria for, 266, 269(Ex. 1), 270(Ex. 10) 

determined by frames, 363 
Local minimization of arc length, 352-358 

M 

Manifold, 184-187 
Mapping 

of Euclidean spaces, 32-41 

of surfaces, 158-166 
Mean curvature, 203, 205-208, 212, 217, 

252 
Mercator projection, 271 (Ex. 13) 
Metric tensor, 305 
Minimal surface, 207, 275(Ex. 5) 

examples, 238 

flat, 263(Ex. 1) 

Gauss mapping, 294(Ex. 6), 296(Ex. 20) 

of revolution, 236-238 

ruled, 233 (Ex. 22) 
Minimization of arc length, 340 
Mobius band, 178, 180(Ex. 7) 
Monge patch, 127, 219 
Monkey saddle, 132(Ex. 6), 205, 231 (Ex. 
9) 

Gaussian curvature, 220(Ex. 5) 

N 

Natural coordinate functions, 4 
Natural frame field, 9 
Neighborhood, 43, 125 

normal, 341 
Norm, 43 
Normal curvature, 195-202, 221 (Ex. 14) 

sign of, 198 
Normal section, 197 
Normal vector field, 147 
n-polygon, 388-389 



O 



1 -segment, 167 
One-to-one, 2 



INDEX 



409 



Onto, 2 

Open interval, 15 

Open set, 5, 43, 151-152 

Orientable surface, 177-178, 185(Ex. 1) 

Orientation 

determined by a patch, 374 

of a frame, 107 

of a patch, 285 

of a paving, 285 

of a surface, 195, 284 

of tangent frame fields, 291 
Orientation-preserving (-reversing) isom- 

etry, 109, 296 
Orientation-preserving (-reversing) re- 

parametrization, 52 
Oriented angle, 291 
Orthogonal coordinates, 276-280 

Gaussian curvature formula, 278 
Orthogonal matrix, 46 
Orthogonal transformation, 100 
Orthogonal vectors, 44 
Orthonormal expansion, 45-46, 83 
Orthonormal frame, see Frame 
Osculating circle, 64(Ex. 6) 
Osculating plane, 61 



Parallel curves, 116 
Parallel postulate, 336 
Parallel surfaces, 209-210 
Parallel translation, 323 
Parallel vector field, 54-55, 322 
Parallel vectors, 6 
Parameter curves, 133 
Parametrization of a curve, 21 
Parametrization in a surface, 135 

criteria for regularity, 136, 140(Ex. 2) 

decomposable into patches, 165(Ex. 9) 
Partial velocities, 134, 136, 146 
Patch, 124-137 

abstract, 182, 184 

geometric computations, 210-216 

Monge, 127 

orthogonal, 220 (Ex. 9), 276-280 

principal, 220(Ex. 9), 280(Ex. 4) 

proper, 124, 152 (Ex. 14) 
Patchlike 2-segment, 281 
Paving, 283, 377 
Planar point, 205 
Plane in E 3 , 60, 131 (Ex. 2), 228, 256 

identified with E 2 , 126 



Poincare 1 half-plane, 309(Ex. 2) 

Gaussian curvature, 317 (Ex. 1) 

geodesies, 337 (Ex. 6) 

isometric to hyperbolic plane, 371 (Ex. 
15) 

polar circles, 351 (Ex. 2) 
Point of application, 6 
Pointwise principle, 8 
Polar circle, 346. 359-360 
Polar disc, 361 (Ex. 2) 
Polygonal region, 386 
Pre-geodesic, 330 
Principal curvatures, 199-207 

as characteristic values, 200 

formula for, 206 
Principal curve, 223-225, 230-232, 263 

(Ex. 4) 
Principal direction, 199 
Principal frame field, 254 
Principal normal, 57, 66, 69 
Principal vectors, 199, 220(Ex. 10), 223 
(Ex. 22) 

as characteristic vectors, 200 
Projective plane, 182-183 

geodesies, 337 (Ex. 8) 

geometric structure, 317 (Ex. 6) 

homogeneity, 371 (Ex. 11) 

natural mapping (projection), 183 

topological properties, 186(Ex. 2) 
Pseudosphere, see Bugle surface 
Pullback, 163 



Quadratic approximation, 202-204 
Quadric surface, 142, 294 (Ex. 10) 

R 

Rectangular decomposition, 377 
Reflection, 109 
Regular curve, 20 
Regular mapping, 38, 161 
Reparametrization, 18 

monotone, 56 (Ex. 10) 

orientation -preserving (-reversing), 52 

unit -speed, 51 
Riemann, 304, 336 
Riemannian geometry, 308, 389-390 
Riemannian manifold, 308 
Rigid motion, see Isometry of Euclidean 

space 
Rigidity, 301 (Ex. 1) 



410 



INDEX 



Rotation, 111 (Ex. 4) 

Ruled surface, 140-143, 227, 231-233 
noncylindrical, 232(Ex. 14) 
total Gaussian curvature, 294 (Ex. 9) 

Ruling, 140 



Saddle surface, 192 
doubly ruled, 227 
Euclidean symmetries, 303 (Ex. 9) 
patch computations, 214-216 
principal vectors, 221 (Ex. 11) 

Scalar multiplication, 3, 8-9 

Scale factor, 268 

Scherk's surface, 222(Ex. 21) 
patch in, 303 (Ex. 11) 
Gauss mapping, 296 (Ex. 20) 

Schwarz inequality, 44 

Serret, 81 

Shape operator, 190-194 
characteristic polynomial, 208 (Ex. 4) 
co variant derivatives and, 324 (Ex. 3) 
as derivative of Gauss mapping, 289 
frame fields, in terms of, 248 
Gaussian and mean curvature, and, 203 
of an immersed surface, 368 
normal curvature and, 196 
preserved by Euclidean isometries, 

297-298 
principal curvatures and vectors, and, 

200 
proof of symmetry, 212-213, 251 (Ex. 6) 

Shortest curve segment, 340 

Sign of an isometry, 108 

Simple region, 294-295 

Simply connected surface, 176, 363 

Slant of a geodesic, 331, 338 

Smooth overlap, 145, 182 

Speed, 51 

Sphere, 128 

conjugate points, 354, 357-358 
Euclidean symmetries, 302(Ex. 8) 
frame-homogeneity, 370 (Ex. 6) 
Gaussian curvature, 207, 219(Ex. 1), 

253, 360, 361 (Ex. 2) 
geodesies, 228-229, 346-347 
geographical patch, 134-135, 277-278 
geometric characterizations, 258, 259, 

262 
geometric structures, 383 
with handles, 379 



holonomy, 323-324, 337 (Ex. 4) 

local isometries, 363-365 

rigidity, 301 (Ex. 1) 

shape operator, 191-192 

topological properties, 176-178, 378 
Spherical curve, 63, 65 (Ex. 10) 
Spherical frame field, 83 

adapted to sphere, 248-249, 277-278 

dual and connection forms, 94-95 
Spherical image 

of a curve, 71, 75 (Ex. 11) 

of a surface, see Gauss mapping 
Standard geometric surface, 363-366 
Stereographic plane, 314 
Stereographic projection, 160, 162 

as conformal mapping, 271 (Ex. 14) 
Stereographic sphere, 314, 337 (Ex. 5) 
Stokes' theorem, 170-172, 387 (Ex. 7) 
Straight line, 15, 18, 55, 229(Ex. 1) 

length-minimizing properties, 56(Ex. 
1) 
Striction curve, 232 (Ex. 14) 
Structural equations 

on E 3 , 92-95 

on a surface, 249, 252, 292, 311-312 
Subset, 1 

Support function, 218, 222, 256 
Surface 

abstract, 182-184 

in E 3 , 125, 306, 367 
implicit definition, 127-128 

geometric, 305 

immersed, 368 
Surface of revolution, 129-130, 234-244 

area, 292 (Ex. 2) 

augmented, 133 (Ex. 12) 

of constant curvature, 239-241, 244 
(Ex. 9), 294 (Ex. 7) 

diffeomorphism types, 187(Ex. 8) 

Gaussian curvature, 235, 238, 242, 
243(Ex. 5) 

geodesies, 338(Ex. 13) 

local characterization, 270(Ex. 12) 

meridians and parallels, 130 

natural frame field, 279 

parametrization 
canonical, 238 
special, 143 (Ex. 13) 
usual, 138-139 

principal curvatures, 235 

principal curves, 225, 235 



INDEX 



411 



topological properties, 182(Ex. 14) 
total curvature, 293 (Ex. 5) 
Symmetry equation, 249 



Tangent bundle, 185 
Tangent line, 22(Ex. 9) 
Tangent plane, 146, 150(Ex. 9) 
Tangent space, 7 
Tangent surface, 231 (Ex. 11) 

isometries of, 270(Ex. 5), 301 (Ex. 2) 
Tangent vector 

to E 3 , 6, 14 

to a surface, 146, 183-184 
Theorema egregium, 273-275 
Topological properties, 176-182, 380n 
Toroidal frame field, 84(#x. 4), 95 (Ex. 2), 

250-251 
Torsion, 58, 66 

formula, 69 

sign, 114-115 
Torus of revolution, 139 

Euler-Poincar£ characteristic, 379, 387 
(Ex. 9) 

Gaussian curvature, 204-205, 235-236 

Gauss mapping, 194(Ex. 5), 290-291 

patch computations, 235-236 

total Gaussian curvature, 287, 291 

usual parametrization, 139 
Total curvature of a curve, 7Q(Ex. 16) 
Total Gaussian curvature, 286-291, 380- 
385 

Euler-Poincare 1 characteristic, and, 380 

Gauss mapping, and, 290 

holonomy and, 325 (.Re. 5) 

of a patch, 325(Ex. 5) 
Total geodesic curvature, 372-375, 386 
(Ex. 4),389CEr. 15) 



Transferred frame field, 272-273 

Translation, 98-100, 109 

Triangle, 383-385 

Triangle inequality, 347 

Triple scalar product, 48-49, 108 

Tube, 221 (Ex. 16) 

2-segment, 169 

U 

Umbilic point, 200, 221 (Ex. 13), see also 

All-umbilic surface 
Unit normal function, 211, 368 
Unit normal vector field, l80(Ex. 5), 190 
Unit points, 34 
Unit speed curve, 51 
Unit sphere, 125 
Unit tangent, 56, 66, 69 
Unit vector, 44 



Vector, see Tangent vector 
Vector analysis, 31 (Ex. 8) 
Vector field 

on an abstract surface, 183 

on a curve, 52-54, 320 

on Euclidean space, 8 

on a surface in E 3 , 147-149, 151 (Ex. 12) 
normal, 147, 149 
tangent, 147, 149 
Vector part, 6 
Velocity, 17-18, 183-184 
Vertices, 373-374 

W 

Wedge product, 27-28, 153 
Winding number, 174(Ex. 5) 



Open 
University 
Course 
Units 0-6 



M334 

Differential 

Geometry 



Q C tf 334 PART * THE OPEN UNIVERSITY 

; Mathematics: A Third Level Course 

516. j 

Jr 

DIP 

PARTO GUIDE TO THE COURSE 



9 



DIFFERENTIAL GEOMETRY 



9 



THE OPEN UNIVERSITY 

Mathematics: A Third Level Course 

DIFFERENTIAL GEOMETRY PART 



O.U. COURSE UNITS COLLECTION 

FOR REFEu£.«CE ONLY 
■ — GUIDE TO THE COURSE 



Prepared by the Course Team 



THE OPEN UNIVERSITY PRESS 



Course Team 



M334 



Chairman: 



Mr. P.E.D. Strain 
Dr. R.A. Bailey 



Lecturer in Mathematics 
Course Assistant in Mathematics 



With assistance from: 



Dr.J.M. Aldous 
Mr. G.J. Burt 
Dr. P.M. Clark 
Mr. P.B. Cox 
Dr. F.C. Holroyd 
Mr. T.C. Lister 
Mr. R.J. Margolis 
Dr. C.A. Rowley 
Mr. M.G. Simpson 



Senior Lecturer in Mathematics 
Lecturer in Educational Technology 
Lecturer in Physics 
Student Computing Service 
Lecturer in Mathematics 
Staff Tutor in Mathematics 
Staff Tutor in Mathematics 
Course Assistant in Mathematics 
Course Assistant in Mathematics 



Consultants: 



Prof. S. Robertson 
Prof. T. Willmore 



Professor of Pure Mathematics, 
University of Southampton 

Professor of Pure Mathematics, 
University of Durham 






.. ;•; b°\ 



3 



;LASSN ^I6^^ "^^ 




/' 



LOAN CAT Kf\^ ""^r 



The Open University Press, Walton Hall, Milton Keynes 

First published 1975 

Copyright © 1975 The Open University 

All rights reserved. No part of this work may be reproduced in any form, by mimeograph or any other means 
without permission in writing from the publishers. 

Produced in Great Britain by 
The Open University Press 

ISBN 335 05700 4 

This text forms part of the correspondence element of an Open University Third Level Course. The complete 
list of parts in the course is given at the end of this text. 

For general availability of supporting material referred to in this text, please write to the Director of 
Marketing, The Open University, P.O. Box 81, Milton Keynes, MK7 6AT. 

Further information on Open University courses may be obtained from The Admissions Office, The Open 
University, P.O. Box 48, Milton Keynes, MK7 6AB. 

1.1 



M334 



CONTENTS Page 

A Brief History of Differential Geometry 5 

Bibliography and Conventions 6 

Prerequisites 7 

Prerequisite Sections 7 

Detailed Prerequisites 8 

Handbooks 10 

The Structure of the Course 11 

1. Breakdown into Blocks 11 

2. Structure of Collection of Blocks 12 

3. Detailed Structure of Blocks and Interconnections 12 

A possible 16-week Breakdown of the Course 16 

Working Instructions 1 7 

Further Reading 2 1 

Errata 22 



M334 



A BRIEF HISTORY OF DIFFERENTIAL GEOMETRY 

This is a very brief history of those aspects of differential geometry that are dealt 
with in this course. It is here to give some historical perspective in a course which 
developes the modern approach to old questions. Differential geometry is a vast 
subject and most of the great mathematicians of the last three centuries have made 
some contribution to it. 

Some of the basic problems of differential geometry were already being studied 
before the introduction of the calculus. For instance, in connection with his study 
of light, Huygens (1629-95) studied the curvature of certain plane curves. However, 
the calculus enabled Newton (1647-1727) to make a more detailed study of such 
curves and Euler (1707-83) was able to study curves on more general surfaces. The 
study of surfaces themselves was opened up by the use of patches introduced in a 
restricted way by Monge (1746-1818) and used by Gauss (1777-1855) to describe 
the curvature of a surface in an intrinsic way. This enabled one to answer such 
questions as whether it was possible to make a plane map of the globe that retained 
all the local properties of its geometry. 

It was Riemann (1826-66) who saw how the theory of surfaces could be generalized. 
The study of Riemann surfaces showed finally that Euclidean geometry was not the 
only possibility. There is now a vast literature on such geometries. 

Meanwhile, the theory of curves took a great step forward with the discovery, by 
Frenet (1816-1900) and Serret (1819-85), of formulas describing the curvature and 
torsion of curves. Their "method of moving frames" was adapted by Darboux 
(1842-1907) to describe surfaces. This method was brought to full generality by 
Cartan (1869-1951), who brought back geometric insight into the profusion of 
equations thrown up by Riemann and his followers. 

A readable introduction to some of the geometric ideas dealt with in this course can 
be found in: 

D. Hilbert and S. Cohn-Vossent: Geometry and the Imagination (Chelsea, 1952). 

This section on differential geometry contains pictures of several of the surfaces we 
shall be studying. 



b M334 

BIBLIOGRAPHY AND CONVENTIONS 
Bibliography 

The set book for this course is Barrett O'Neill: Elementary Differential Geometry 
(Academic Press, 1966). It is essential to have this book: the course is based on it 
and will not make sense without it. 

The set books for M201, M231 and MST 282 are referred to occasionally; they are 
useful but not essential. They are: 

D.L. Kreider, R.G. Kuller, D.R. Ostberg and F.W. Perkins: An Introduction to 
Linear Analysis (Addison-Wesley, 1966). 

E.D. Nering: Linear Algebra and Matrix Theory (John Wiley, 1970). 

M. Spivak: Calculus, paperback edition (W.A. Benjamin/ Addision-Wesley, 1973). 

R.C. Smith and P. Smith: Mechanics, SI Edition (John Wiley, 1972). 



Conventions 

Unreferenced pages and sections denote the set book. Otherwise 

O'Neill denotes the set book; 

Text denotes the correspondence text; 

KKOP denotes An Introduction to Linear Analysis by D.L. Kreider, 
R.G. Kuller, D.R. Ostberg and F.W. Perkins; 

Nering denotes Linear Algebra and Matrix Theory by E.D. Nering; 

Spivak denotes Calculus by M. Spivak; 

Smith denotes Mechanics by R.C. Smith and P. Smith. 

Reference to Open University Courses in Mathematics take the form: 
Unit Ml 00 22, Linear A Igebra I 
Unit MST 281 10, Taylor Approximation 
Unit M2 01 16, Euclidean Spaces I: Inner Products 
Unit M231 2, Functions and Graphs 
Unit MST 282 1, Some Basic Tools. 

References to Definitions, Theorems, Lemmas etc. in the set book, O'Neill, are given 
as in the following example : 

Lemma V.4.2 stands for Lemma 4.2 of Chapter V. 

Sometimes if O'Neill is referring to something in the same chapter he omits the 
roman numeral for the Chapter. 



M334 



PREREQUISITES 

The courses that are prerequisites for this course are M201 Linear Mathematics, 
M231 Analysis, and MST 282 Mechanics and Applied Calculus. 

However, this course has been written with very few explicit references to any of 
these courses. In this section we list those units and sections of the prerequisite 
courses that we have built on, and this is followed by a brief description of their 
content. This should enable you to work out whether you are familiar with those 
ideas and results that we shall be assuming. The bracketed sections below are marked 
as optional to M231 and so you may not have read them yet. If you want to, do so 
now, noting only the ideas and results. The details of the definitions and proofs will 
not be needed and our notation will not be similar. 



Prerequisite Sections 

M201 Linear Mathematics 

Unit 1 Vector Spaces 

Unit 2 Linear Transformations 

Unit 3 Hermite Normal Form, Section 3.3 

Unit 5 Determinants and Eigenvalues 

Unit 10 Jordan Normal Form, Section 10.2.1 

Unit 12 Linear Functionals and Duality, Section 12.1 

Unit 14 Bilinear and Quadratic Forms, Sections 14.1 and 14.2.3 

Unit 15 Affine Geometry and Convex Cones, Sections 15.1.1 and 15.1.2 

Unit 16 Euclidean Spaces I, Inner Proudcts 

Unit 23 The Wave Equation, Section 23.1.1 

Unit 24 Orthogonal and Symmetric Transformations, Sections 24.1, 24.3.1 and 
24.3.2 

M231 Analysis 

Unit 2 Functions and Graphs (plus Section 2.7) 

Unit 5 Differentiation (plus Section 5.9) 

(Unit 7 Applications of the Derivative, Section 7.8) 

Unit 10 Some Important Functions 

MST 282 Mechanics and Applied Calculus 

Unit 1 Some Basic Tools, Sections 1.1. to 1.3 
Unit 2 Kinematics, Sections 2.1.1 and 2.2 
Unit 7 Work and Energy I, Section 7.1.2 



8 M334 

Detailed Prerequisites 

M201 Linear Mathematics 

This is the major prerequisite for our course. Though we do not often quote results 
from it, we freely use all the basic ideas and conventions that it introduced. 

Unit 1 Vector Spaces This unit introduces the vector space of arrows, R 2 and 

R 3 , which are the basis for our course. You should also know about linear 
dependence, bases and coordinates, though we shall deal only with concrete 
examples in R 3 . Ideas such as dimension and subspaces are implicit in our course but 
you will not need any explicit results. In fact, the ideas of this unit are basic to all 
linear mathematics. 

Unit 2 Linear Transformations This unit again introduces ideas basic to our 

course. You will need to know about linear transformations , the convention for 
representing them by matrices, isomorphisms and rank, though we shall be in- 
terested in 2- and 3-dimensional examples only. This is another unit with which to 
be entirely familiar. 

Unit 3 Hermite Normal Form We do not use the technique of Hermite normal 

forms in our course. However, we do talk about the rank of 2 X 2 and 3X3 
matrices, which can be read off the Hermite normal form obtained by the elemen- 
tary operations in Section 3.3. 

Unit 5 Determinants and Eigenvalues Determinants, eigenvalues and eigen- 
vectors play a small but important part in our course. The cross product of two 
vectors will be defined in terms of 3 X 3 determinants and you should be familiar 
with the formula for expanding such a determinant dealt with in KKOP, pages 
682 - 687, where the effect of column operations is also described. In Section 5.1.5 it 
is shown that row operations behave similarly and it is these that we shall use. You 
will need to know only what eigenvectors and eigenvalues are and not in general how 
to calculate them. We shall be using eigenvector bases as in Section 5.2.4. 

Unit 10 Jordan Normal Form In Section 10.2.1 it is shown that a linear trans- 

formation is represented by a diagonal matrix with respect to an eigenvector basis. 

Unit 12 Linear Functional and Duality It will be useful to know the definition 

of linear functionals, the dual space and a dual basis. However, do not worry about 
any technical results. Since we shall be dealing with R 2 and R 3 , in which there is a 
simple inner (scalar, dot) product, we shall be able to describe dual bases etc. in a 
concrete way. 

Unit 14 Bilinear and Quadratic Forms Again you will find it useful to have met 

bilinear forms and symmetric bilinear forms. It will give you some more insight into 
the course, although the only explicit result we shall use is the quadratic Taylor 
approximation. 

Unit 15 A f fine Geometry and Convex Cones Though this unit is not in the spirit 

of our course, Sections 15.1.1 and 15.1.2 give various ways of describing affine lines 
and planes which may prove useful to you. 



M334 



PREREQUISITES 

The courses that are prerequisites for this course are M201 Linear Mathematics, 
M231 Analysis, and MST 282 Mechanics and Applied Calculus. 

However, this course has been written with very few explicit references to any of 
these courses. In this section we list those units and sections of the prerequisite 
courses that we have built on, and this is followed by a brief description of their 
content. This should enable you to work out whether you are familiar with those 
ideas and results that we shall be assuming. The bracketed sections below are marked 
as optional to M231 and so you may not have read them yet. If you want to, do so 
now, noting only the ideas and results. The details of the definitions and proofs will 
not be needed and our notation will not be similar. 



Prerequisite Sections 

M201 Linear Mathematics 

Unit 1 Vector Spaces 

Unit 2 Linear Transformations 

Unit 3 Hermite Normal Form, Section 3.3 

Unit 5 Determinants and Eigenvalues 

Unit 10 Jordan Normal Form, Section 10.2.1 

Unit 12 Linear Functionals and Duality, Section 12.1 

Unit 14 Bilinear and Quadratic Forms, Sections 14.1 and 14.2.3 

Unit 15 Affine Geometry and Convex Cones, Sections 15.1.1 and 15.1.2 

Unit 16 Euclidean Spaces I, Inner Proudcts 

Unit 23 The Wave Equation, Section 23.1.1 

Unit 24 Orthogonal and Symmetric Transformations, Sections 24.1, 24.3.1 and 
24.3.2 

M231 Analysis 

Unit 2 Functions and Graphs (plus Section 2.7) 

Unit 5 Differentiation (plus Section 5.9) 

(Unit 7 Applications of the Derivative, Section 7.8) 

Unit 10 Some Important Functions 

MST 282 Mechanics and Applied Calculus 

Unit 1 Some Basic Tools, Sections 1.1. to 1.3 
Unit 2 Kinematics, Sections 2.1.1 and 2.2 
Unit 7 Work and Energy I, Section 7.1.2 



8 M334 

Detailed Prerequisites 

M201 Linear Mathematics 

This is the major prerequisite for our course. Though we do not often quote results 
from it, we freely use all the basic ideas and conventions that it introduced. 

Unit 1 Vector Spaces This unit introduces the vector space of arrows, R 2 and 

R 3 , which are the basis for our course. You should also know about linear 
dependence, bases and coordinates, though we shall deal only with concrete 
examples in R 3 . Ideas such as dimension and subspaces are implicit in our course but 
you will not need any explicit results. In fact, the ideas of this unit are basic to all 
linear mathematics. 

Unit 2 Linear Transformations This unit again introduces ideas basic to our 

course. You will need to know about linear transformations, the convention for 
representing them by matrices, isomorphisms and rank, though we shall be in- 
terested in 2- and 3-dimensional examples only. This is another unit with which to 
be entirely familiar. 

Unit 3 Hermite Normal Form We do not use the technique of Hermite normal 

forms in our course. However, we do talk about the rank of 2 X 2 and 3X3 
matrices, which can be read off the Hermite normal form obtained by the elemen- 
tary operations in Section 3.3. 

Unit 5 Determinants and Eigenvalues Determinants, eigenvalues and eigen- 
vectors play a small but important part in our course. The cross product of two 
vectors will be defined in terms of 3 X 3 determinants and you should be familiar 
with the formula for expanding such a determinant dealt with in KKOP, pages 
682 _ 687, where the effect of column operations is also described. In Section 5.1.5 it 
is shown that row operations behave similarly and it is these that we shall use. You 
will need to know only what eigenvectors and eigenvalues are and not in general how 
to calculate them. We shall be using eigenvector bases as in Section 5.2.4. 

Unit 10 Jordan Normal Form In Section 10.2.1 it is shown that a linear trans- 

formation is represented by a diagonal matrix with respect to an eigenvector basis. 

Unit 12 Linear Functional and Duality It will be useful to know the definition 

of linear functionals, the dual space and a dual basis. However, do not worry about 
any technical results. Since we shall be dealing with R 2 and R 3 , in which there is a 
simple inner (scalar, dot) product, we shall be able to describe dual bases etc. in a 
concrete way. 

Unit 14 Bilinear and Quadratic Forms Again you will find it useful to have met 

bilinear forms and symmetric bilinear forms. It will give you some more insight into 
the course, although the only explicit result we shall use is the quadratic Taylor 
approximation. 

Unit 15 A f fine Geometry and Convex Cones Though this unit is not in the spirit 

of our course, Sections 15.1.1 and 15.1.2 give various ways of describing affine lines 
and planes which may prove useful to you. 



M334 9 

Unit 16 Euclidean Spaces I: Inner Products Sections 16.1 to 16.3 are very 

important to our course. They introduce the inner (scalar, dot) product of vectors, 
which we use extensively. You will need to be able to use it to define length, 
distance, angle and orthonormal bases. We shall frequently use orthonormal bases, 
orthonormal expansions and the simple form of the inner product in terms of the 
coordinates with respect to such a basis, all of which are dealt with in Section 
16.4.3. Our course goes over all the details again. 

Unit 23 The Wave Equation All you need from this unit is the definition of 

partial differention in Section 23.1.1, though even this will be redefined. 

Unit 24 Orthogonal and Symmetric Transformations This unit contains many 

results that are needed in this course. Orthogonal transformations are linear space 
'morphisms' and the orthogonal matrices representing them are characterized in 
several important ways, including the result that their inverse is equal to their 
transpose. We shall also use the result that symmetric transformations can be repre- 
sented by diagonal matrices with respect to suitable eigenvector bases. 



M231 Analysis 

Unit 2 Functions and Graphs This unit covers all the basic ideas associated with 

graphs and functions of one variable with which you need to be familiar. 

The appendix, Section 2.7, is very straightforward and you ought to have a look at 
it. It introduces ideas that will be developed in this course. 

Unit 5 Differentiation This unit introduces the basic definitions of differentia- 

tion and curves. These are ideas which will be developed further in this course. 
However, our approach will be less theoretical. We shall be interested in using 
techniques rather than worrying about such problems as differentiability. 

We shall also be using partial and directional derivatives extensively. These are intro- 
duced in an appendix, Section 5.9. If you have looked at this before you might like 
to look at it again, otherwise don't worry. 

Unit 7 Applications of the Derivative The only relevant part of this unit is the 

appendix, Section 7.8, on implicit functions. This is straightforward and will help 
you with Chapter IV of this course. Try to look through it before then but do not 
bother with any proofs or the sections on contunuity and differentiability. 

Unit 10 Some Important Functions This unit introduces all the standard func- 

tions that you will need in this course. You will need to know about polynomials, 
trigonometric functions, the exponential and logarthmic functions, hyperbolic 
functions and inverse trigonometric and hyperbolic functions. Useful relations in- 
volving these and their derivatives can be found in the summary, Section 10.10. 



MST 282 Mechanics and Applied Calculus 

Unit 1 Some Basic Tools Sections 1.1 to 1.3 give a useful introduction to 

vectors and applied calculus, with which you should be familiar. We shall use the 
scalar, vector and triple scalar products of vectors and some simple forms of the 
chain rule for differentiation. 



10 M334 

Unit 2 Kinematics In Section 2.1.1 differentiation of a vector is introduced and 

in Section 2.2 parametric equations of curves are discussed. These are ideas that will 
be developed further and so you should be familiar with them. 

Unit- 7 Work and Energy I Section 7.1.2 contains some very important ideas 

that we shall develop further and so you should be familiar with them. It deals with 
scalar fields, level curves and surfaces, directional derivatives and the gradient. It also 
contains a simple form of the chain rule for differentiation, for which it gives a 
proof in Appendix 1. 



Handbooks 

Each of the prerequisite courses has a handbook which contains some of the defini- 
tions to which we refer. However, the most useful parts of the handbooks are those 
giving lists of the derivatives of standard functions. These are 

M231 Handbook: Section 5, Familiar Functions, pages 45-48 
MST 282 Handbook: Section 4, Table of Derivatives, page 27. 



M334 



11 



THE STRUCTURE OF THE COURSE 

This course is based on B. O'Neill's book Elementary Differential Geometry pub- 
lished by Academic Press. Again we must stress that it is essential to have this book 
since the course is based almost entirely on it. The course consists of the first six 
chapters of the book, apart from a few odd sections, and, with one minor exception, 
you should read the book in the order it is written. We expect you to read it section 
by section and we have provided a commentary on each relevant section. The 
commentaries have been bound together chapter by chapter. 

There is another way of viewing the structure of the course which might prove 
useful. All those sections dealing with a particular theme can be thought of as 
forming a block. We shall show which sets of sections form blocks and also how 
these blocks are interrelated, that is, which blocks it is essential to have tackled, at 
least in part, before going onto another block. If you ever become stuck with the 
sections in one block this will enable you to see if there is any alternative block that 
you can tackle first. 

Sometimes a particular section does not depend on all those sections that have 
preceded it in that block or on all the sections of preceding blocks. We have included 
this information in the form of a detailed structure diagram for each block giving all 
prerequisite blocks and sections. 

Each introduction to a section again lists the most important prerequisite sections 
but we hope that if you refer to the structure diagrams below regularly the overall 
structure of the course will become clearer. 



1. Breakdown into Blocks 

Here is the content of the blocks and the number of units to which they are 
equivalent, assuming that this is a standard x h credit course with fourteen teaching 
units and two gaps. 



B. 



D. 



E. 



H. 



Calculus and Linear Algebra in E 3 
1.1 to 1.4, 1.7, II. 1, II.2, II.5, II.6 

Basic Exterior Calculus in E 3 
1.5 and 1.6 

Connection Forms for Frame Fields in E 3 
II. 7 and II. 8 

Frenet Formulas for Curves in E 3 
II. 3 and II.4 

Geometry of E 3 
III.l to III. 3 

Geometry of Curves in E 3 
III.4 and III.5 

Calculus on Surfaces in E 3 
1V.1 to IV.3, IV.5(a) 

Exterior Calculus on Surfaces in E 3 
IV.4, IV.5(b), IV.7(a) 



3 units 
x h unit 
x h unit 
1 unit 
x h unit 
x h unit 
\ x /z units 
1 unit 



12 M334 

I. Shape Operator of a Surface in E 3 

V.ltoV.4 2 units 

J. Applications of the Shape Operator 

V.5 and V.6 1 unit 

K. Structure of Surfaces in E 3 

VI. 1, VI.2,VL6 1 unit 

L. Global results on Surfaces in E 3 

VI.3, IV.7(b) Vi unit 

M. Geometry of Surfaces in E 3 

VI.4, VI.5, VI.8 1 unit 



2. Structure of Collection of Blocks 

The blocks are related as below, a block is preceded by all those blocks above it to 
which it is connected. 




So, for instance, even if you find block C too difficult, and on first reading it is quite 
difficult, there is no reason why you should not press on with D, E, F, G, H, I, J as 
long as you have looked at C again before you come to K that depends on it. 

Each block and all those that precede it tell an interesting story. For instance the 
line terminating at J consists of the blocks A, D, G, I, J and describes the shape of 
surfaces of revolution and interesting curves on surfaces. 



3. Detailed Structure of Blocks and Interconnections 

In the following diagrams sections in the same block are grouped together. Arrows 
indicate how the sections are interconnected. We have shown only how a section 
depends on those directly preceding it. They in turn may depend on others. 



M334 



13 




1.1 

i 

1.2 

1 

1.3 

1 

1.4 

1 

1.7 



B 



II.2 



II.6 



II.5 



A 

1.3 



1.5 

1 

1.6 
B 




D 



A 

II.2 



II.3 

1 

II.4 
D 



II. 1 


A 


1.7 












II] 


1 _ 




' 


' 




.1 




11 
I 


1.2 


* 111. J 



D 



II.3 



II.4 



E 

III.3 

* III.4 

1 

^ III.5 



14 



M334 




H 




G 

IV.3 




D 

II.3 




















' 


' 




^ ' 








V 


.1 




■*• V 


.z 


I 


* v .5 


w V .4 



M334 



15 






M 







G 

IV.5 


l 


H 

IV.5 




K 

VI. 1 




F 

III.5 














\ 




/ 








J 


V.6- 




i 


- 








' 


' 




Vl-4 


*■ \ 


► 


\n q 








i 


M 




















III.2 
E 







16 



M334 



A POSSIBLE 16-WEEK BREAKDOWN OF THE COURSE 

In case you wish to break down the course into 16 weeks of work, the following is 
a suggested way of doing so. 



Week 


Sections of O'Neill 


Blocks 


TMA 


1 


1.1,1.2,1.3,1.4 


A 






2 


1.5,1.6,1.7 


A, B 


01 




3 


11.1,11.2 


A 






4 


11.3,11.4 


D 






5 


11.5,11.6,11.7,11.8 


A,C 






6 






02 




7 


III.l, IIL2, III.3, III.4, III.5 


E,F 






8 


IV.1,IV.2 


G 






9 


IV.3,IV.4 


G,H 


03 




10 


IV.5, IV.7(a) 


G,H 






11 


V.1,V.2,V.3 


I 






12 


V.4 


I 






13 


V.5, V.6 


J 


04 




14 


VI.l,VI.2,VI.3,IV.7(b) 


K, L 






15 


VI.4,VI.5,VI.6 


K,M 






16 


VI. 8 


M 







M334 17 

WORKING INSTRUCTIONS 

This section deals with instructions on how to tackle written material of this course. 
Of course, we realize that by now that you will probably have worked out which 
way it is best for you to tackle an Open University text and you may not wish to 
change successful habits. However, while we were writing our commentaries we had 
to keep in mind a certain approach to reading them and this was responsible for 
them being structured in the way they are. 



Components 

This course has two components. The first is the Set Book, B. O'Neill: Elementary 
Differential Geometry, and the second is the Course Notes. The set book is divided 
into Chapters which correspond to Parts of the course notes and both of these are 
divided into Sections indexed in the same way. 



Reading a Section 

Each section is divided up in the following way. 

Introduction 

Each section begins with a short introduction. The content of this introduction 
depends on just how much of an introduction there is in the set book. The intro- 
duction tells you the prerequisite sections and possibly some essential results that 
will be used. It goes on to describe very briefly the content of the section, in terms 
of those that have gone before, and how this section fits into the overall structure of 
the course. It will also tell you if it is possible to put this section off till some later 
time. 

Reading Passage 

This will usually be an entire section of the set book and it should be read after the 
introduction. We expect you to read it more than once. 

First Reading At the first reading you should try to browse through all the 

reading passage and obtain an overall view. Look carefully at each definition and try 
to think of examples. This is a very geometric course so you should try to draw lots 
of pictures for yourself. Look carefully at any Theorems or Lemmas, making sure 
you understand all the terms on both sides of each equation, but at this reading do 
not worry too much about detailed proofs. You should also try and follow the steps 
in any straightforward worked examples. There are several manipulative techniques 
that are demonstrated in this way and you will need to master them. 



18 M334 

Comments 

After completing the reading passage you should next look at the comments we have 
provided. Though they are related to the text you are not expected to jump back- 
wards and forwards from the reading page to them. In fact some of the comments 
are quite general and relate to an idea that is developed throughout the section 
rather than some specific result. You should read these comments in the same spirit 
as the reading passage. 



Additional Text 

The Additional Text should be read directly after the Reading Passage and the 
Comments and approached in exactly the same way. The Additional Text consists of 
definitions and results that we feel should be discussed at this point and it should be 
treated as an integral part of the section. 



Optional Text 

Some sections will contain an Optional Text. These consist either of passages of the 
text book or of the course notes and develop certain straightforward topics that we 
consider interesting but peripheral to the main structure of the course. We have 
included these Optional Texts for those of you who are finding the course enticing 
and have time'to spare; others may omit them without losing anything. 

Text Exercises 

At the end of the Comments, Additional or Optional Texts you may find some Text 
Exercises. These are included at this point because they form an integral part of the 
text. They may deal with some result that has already been referred to or with some 
result that will be of crucial importance later on. 

Second Reading After having worked once through the Reading Passage, Com- 

ments and Additional Text you should work through them again but this time in 
more detail. Once you have grasped the overall structure of the section you need to 
fill in the finer details. Look up references needed to complete a proof and work 
through all proofs and examples filling in the missing steps. 



Supplementary Comments 

It is not essential that you read these comments. They are intended only to help you 
at the second reading. They are intended to cover those tricky or routine points 
which are not essential to the development of the subject but may nevertheless cause 
some concern. Anything not covered by either the Comments or the Supplementary 
Comments has been omitted because we do not consider it of any importance. 

Comments Versus Supplementary Comments The main difference between 

these two types of comments is that Comments are an integral part of the text and 
should be included at each reading and during revision, while Supplementary 
Comments need be looked at only once to clear up tricky points and are not 
essential to revision. 



M334 17 

WORKING INSTRUCTIONS 



This section deals with instructions on how to tackle written material of this course. 
Of course, we realize that by now that you will probably have worked out which 
way it is best for you to tackle an Open University text and you may not wish to 
change successful habits. However, while we were writing our commentaries we had 
to keep in mind a certain approach to reading them and this was responsible for 
them being structured in the way they are. 



Components 

This course has two components. The first is the Set Book, B. O'Neill: Elementary 
Differential Geometry, and the second is the Course Notes. The set book is divided 
into Chapters which correspond to Parts of the course notes and both of these are 
divided into Sections indexed in the same way. 



Reading a Section 

Each section is divided up in the following way. 

Introduction 

Each section begins with a short introduction. The content of this introduction 
depends on just how much of an introduction there is in the set book. The intro- 
duction tells you the prerequisite sections and possibly some essential results that 
will be used. It goes on to describe very briefly the content of the section, in terms 
of those that have gone before, and how this section fits into the overall structure of 
the course. It will also tell you if it is possible to put this section off till some later 
time. 

Reading Passage 

This will usually be an entire section of the set book and it should be read after the 
introduction. We expect you to read it more than once. 

First Reading At the first reading you should try to browse through all the 

reading passage and obtain an overall view. Look carefully at each definition and try 
to think of examples. This is a very geometric course so you should try to draw lots 
of pictures for yourself. Look carefully at any Theorems or Lemmas, making sure 
you understand all the terms on both sides of each equation, but at this reading do 
not worry too much about detailed proofs. You should also try and follow the steps 
in any straightforward worked examples. There are several manipulative techniques 
that are demonstrated in this way and you will need to master them. 



1Q M334 

lo 

Comments 

After completing the reading passage you should next look at the comments we have 
provided. Though they are related to the text you are not expected to jump back- 
wards and forwards from the reading page to them. In fact some of the comments 
are quite general and relate to an idea that is developed throughout the section 
rather than some specific result. You should read these comments in the same spirit 
as the reading passage. 



Additional Text 

The Additional Text should be read directly after the Reading Passage and the 
Comments and approached in exactly the same way. The Additional Text consists of 
definitions and results that we feel should be discussed at this point and it should be 
treated as an integral part of the section. 



Optional Text 

Some sections will contain an Optional Text. These consist either of passages of the 
text book or of the course notes and develop certain straightforward topics that we 
consider interesting but peripheral to the main structure of the course. We have 
included these Optional Texts for those of you who are finding the course enticing 
and have time to spare; others may omit them without losing anything. 

Text Exercises 

At the end of the Comments, Additional or Optional Texts you may find some Text 
Exercises. These are included at this point because they form an integral part of the 
text. They may deal with some result that has already been referred to or with some 
result that will be of crucial importance later on. 

Second Reading After having worked once through the Reading Passage, Com- 

ments and Additional Text you should work through them again but this time in 
more detail. Once you have grasped the overall structure of the section you need to 
fill in the finer details. Look up references needed to complete a proof and work 
through all proofs and examples filling in the missing steps. 



Supplementary Comments 

It is not essential that you read these comments. They are intended only to help you 
at the second reading. They are intended to cover those tricky or routine points 
which are not essential to the development of the subject but may nevertheless cause 
some concern. Anything not covered by either the Comments or the Supplementary 
Comments has been omitted because we do not consider it of any importance. 

Comments Versus Supplementary Comments The main difference between 

these two types of comments is that Comments are an integral part of the text and 
should be included at each reading and during revision, while Supplementary 
Comments need be looked at only once to clear up tricky points and are not 
essential to revision. 



M334 19 

Summary 

After mastering as much of the Reading Passage, Comments and Additional Text as 
you can, you should look through the Summary provided in the Course Notes. Of 
course if you feelllike preparing your own summary and comparing it with ours you 
might find it a worthwile exercise. Our summary has several subsections. First comes 
Notation, then Definitions, Results, Examples and Techniques. 



Notation We have included only representative examples of each notation intro- 

duced in this section. 



Definitions and Results We have included in our list of Definitions and Results 

some that O'Neill has mentioned only in passing but which we feel are important 
enough to be considered on a par with all the others. 

Examples We have included only those examples that are dealt with regularly 

throughout the course. Important results about these examples are recorded in later 
Results subsections. 



Techniques These techniques could be referred to as the aims of the Sections. 

For some of the techniques you need to be able to recognize some new object. For 
others you need to be able to perform some straightforward manipulation, while for 
others you need to be able to apply some abstract result or reproduce an argument 
similar to one used in the text. 



Exercises 

After inspecting the Summary you should try the Exercises without looking at the 
solutions. The Exercises fall into the following categories. 



Text Exercises These were mentioned above. 



Technique Exercises These are the most important exercises and are the type we 

shall expect you to have mastered. We have related each exercise to the relevant 
techniques so that you know exactly how you ought to tackle it. 



Theory Exercises There will be only a few of this type of exercise. They derive 

some abstract result or slightly extend those dealt with in the Section. 



Optional Exercises These are based on or sometimes constitute the Optional 

Text and so are only for those working through the Optional Texts. 



Solutions 

The solutions follow the exercises. We have tried to give fairly lengthy solutions to 
the early exercises, pointing out the reasons for all the steps, but as time goes on we 
omit reasoning that has been used previously. 



20 M334 

Further Exercises 

At the end of our commentary on each Chapter there will be a selection of Further 
Exercises that can be used for self-assessment or revision purposes. They will be 
either Technique Exercises or Other Recommended Exercises. The techniques 
needed will be those of the relevant section and each of the other Recommended 
Exercises will be accompanied by a brief description of how the exercise fits into the 
course. The Further Exercises will be followed by very brief Solutions. 



M334 

FURTHER READING 

If you are interested in learning more about the subject after you have completed 
the course, then there are the remaining sections for you to read. 

Sections IV.6 and VI. 7 deal with integration on a surface in E 3 ; Section IV.8 
introduces manifolds, and Chapter VII applies all the techniques and results of the 
preceding Chapters to the study of Riemannian Geometry. 

O'Neill also provides a short bibliography on page 391. For more differential geo- 
metry, at about the same level, you should read the book by T.J. Willmore; for a 
more advanced approach you should read the book by N. J Hicks, and for applica- 
tions the book by H. Flanders. 

There are a large number of other books now available on differential geometry and 
it is also the subject of many current research papers. We hope that this course serves 
you as an introduction to this literature. 



22 M334 

ERRATA 

The following list comprises the significant mathematical errors we have found in 
the text of O'Neill. It does not include typographical errors or mistakes where the 
meaning is nonetheless clear. Errors in exercises and their solutions are generally 
given not here but where the exercises are mentioned in the correspondence text. 

Chapter I 

Page 25, last line of text: The coefficient of v 3 should be {p\ + 2). 
Page 31, Exercise 6: 'V in the last line should be "#". 
Page 37, line 16: "Corollary 7.6" should be "Theorem 7.5". 



d 2 a 2 



Chapter II 

Page 54, line 9: The second term of a" should be 

dt 2 

Page 80, Exercise 1: "VyW" should be "V V W". 

Chapter III 

Page 102, line -2: "q = F(q)" should be "q = F(p)". 
Page 108, line -8: This should read 

Page 108, lines -5, -4: The right-hand side of these should read 

" = det(,4*C) 
= det ^4.det *C = det ^4.det C". 

Page 120, line -7: The last two "F"s should be removed. 

Oiapter IV 

Page 136, line -5: The final term should be sinV 
Page 148, Lemma 3.8: This should read 

"If M: g = c is a surface in E 3 and if the gradient vector field 
Vg= X (3#/d*Y) U( (considered only at points of M) is never zero, then it 
is a non-vanishing normal vector field on the entire surface M." 

The first three lines of the proof should be omitted. 

Page 157, Exercise 5. The right-hand side of the first equation should be 

S'(f(p))v p [/1. 



M334 23 

Chapter V 

Page 204, lines 9 and -2: "T p {M)" should be "Tp(M)" in each case. 
Page 210, line -9: The final "V should be "x„". 



Page 214, line -6: "-1" should be "-1/6 



2» 



Page 218, line -4 and diagram: "7p(iW)» should be "Tp{M)" in each case. 
Page 241, line -5: The value of each curvature should be multiplied by -1. 



Chapter VI 

Page 252, line 18: The matrix should be transposed. 
Page 297, line 6 of text: The second surface should be 

M: z = * 2 -y 2 . 
2 

Page 299, line -2: The right-hand side of the equation should be 



24 M334 

DIFFERENTIAL GEOMETRY 

I Calculus on Euclidean Space 

II Frame Fields 

III Euclidean Geometry 

IV Calculus on a Surface 

V Shape Operators 

VI Geometry of Surfaces in E 3 



WpW* 



PART I THE OPEN UNIVERSITY 



k<\c Mathematics: A Third Level Course ^^ 

DIP 

PART I CALCULUS ON EUCLIDEAN SPACE 



DIFFERENTIAL GEOMETRY 



9 




THE OPEN UNIVERSITY 
Mathematics: A Third Level Course 



D [FFERENTIAL GEOMETRY PART I 



FOR REf ek:.;ce only 



CALCULUS ON EUCLIDEAN SPACE 



A commentary on Chapter I of O'Neill's 

Elementary Differential Geometry 

Prepared by the Course Team 



THE OPEN UNIVERSITY PRESS 



Course Team 



M334I 



Chairman: 



Mr. P.E.D. Strain 
Dr. R.A. Bailey 



Lecturer in Mathematics 
Course Assistant in Mathematics 



With assistance from: 



Dr. J.M. Aldous 
Mr. G.J. Burt 
Dr. P.M. Clark 
Mr. P.B. Cox 
Dr. F.C. Holroyd 
Mr. T.C. Lister 
Mr. R.J. Margolis 
Dr. C.A. Rowley 
Mr. M.G. Simpson 



Senior Lecturer in Mathematics 
Lecturer in Educational Technology 
Lecturer in Physics 
Student Computing Service 
Lecturer in Mathematics 
Staff Tutor in Mathematics 
Staff Tutor in Mathematics 
Course Assistant in Mathematics 
Course Assistant in Mathematics 



Consultants: 



Prof. S. Robertson 



Prof. T. Willmore 



Professor of Pure Mathematics, 
University of Southampton 

Professor of Pure Mathematics, 
University of Durham 



SAK CGUc Ha 



*G3 



CLASS No. 





The Open University Press, Walton Hall, Milton Keynes 

First published 1975 

Copyright © 1975 The Open University 

All rights reserved. No part of this work may be reproduced in any form, by mimeograph or any other means, 
without permission in writing from the publishers. 

Produced in Great Britain by 
The Open University Press 

ISBN 335 05700 4 

This text forms part of the correspondence element of an Open University Third Level Course. The complete list 
of parts in the course is given at the end of this text. 

For general availability of supporting material referred to in this text, please write to the Director of Marketing, 
The Open University, P.O. Box 81, Milton Keynes, MK7 6AT. 

Further information on Open University courses may be obtained from The Admissions Office, The Open 

University, P.O. Box 48, Milton Keynes, MK7 6AB. 

1.1 



M334I 



CONTENTS Page 

Set Book 4 

Bibliography 4 

Conventions 4 

1.1 Introduction and Euclidean Space 5 

1.2 Tangent Vectors 10 

1.3 Directional Derivatives 16 

1.4 Curves in E 3 22 

1.5 1-Forms 30 

1.6 Differential Forms 37 

1.7 Mappings 43 
Further Exercises and Solutions 54 



4 M334 1 

Set Book 

Barrett O'Neill: Elementary Differential Geometry (Academic Press, 1966). It is 
essential to have this book: the course is based on it and will not make sense without 
it. 



Bibliography 

The set books for M201, M231 and MST 282 are referred to occasionally; they are 
useful but not essential. They are: 

D.L. Kreider, R.G. Kuller, D.R. Ostberg and F.W. Perkins: An Introduction to 
Linear Analysis (Addison-Wesley, 1966). 

E.D. Nering: Linear Algebra and Matrix Theory (John Wiley, 1970). 

M. Spivak: Calculus, paperback edition (W.A. Benjamin/ Addision-Wesley, 1973). 

R.C. Smith and P. Smith: Mechanics, SI Edition (John Wiley, 1972). 



Conventions 

Before starting work on this text, please read M334 Part Zero. Consult the Errata 
List and the Stop Press and make any necessary alterations for this chapter in the set 
book. 

Unreferenced pages and sections denote the set book. Otherwise 

O'Neill denotes the set book; 

Text denotes the correspondence text; 

KKOP denotes An Introduction to Linear Analysis by D.L. Kreider, R.G. 
Kuller, D.R. Ostberg and F.W. Perkins; 

Nering denotes Linear Algebra and Matrix Theory by E.D. Nering; 

Spivak denotes Calculus by M. Spivak; 

Smith denotes Mechanics by R.C. Smith and P. Smith. 

Reference to Open University Courses in Mathematics take the form: 
Unit Ml 00 22, Linear Algebra I 
Unit MST 281 10, Taylor Approximation 
Unit M201 16, Euclidean Spaces I: Inner Products 
Unit M231 2, Functions and Graphs 
Unit MST 282 1, Some Basic Tools. 



M334I.1 5 

1.1 INTRODUCTION AND EUCLIDEAN SPACE 

Introduction 

In this section we introduce some notation that will be used throughout the course. 
We review vector spaces and partial differentiation. In the additional text we give 
some rules for the evaluation of partial derivatives. 

READ: Introduction and Section 1.1 (pages 1-5). 

Comments 

(i) Page 1: line -1 In the correspondence text we shall sometimes write 

g o f for the composite function. 

(ii) Page 4: Definition 1.2 From now on we shall use x, y, z in this way 

only. We shall not use x as a typical element of the real line, or indeed any 
set as O'Neill does on pages 1 and 2 of the Introduction. 

Additional Text 

Partial differentiation Suppose p = (p 1} p 2 , P3) is a point of E 3 and f : E 3 ► R 

is a differentiable function. To obtain (3f/dx 1 )(p) we first construct the function 

F : R >R such that F(t) = f(t, p 2 ,p 3 ) and then (3f/8x 1 )(p) =F'(p 1 ). That is we 

consider x 2 and x 3 as constants and differentiate in the usual way. Partial differen- 
tiation satisfies the following rules. Let a, b be real numbers; f, g differentiable func- 
tions from E 3 to R and h a differentiable function from R to R, then 

(1) a <" fH *> = , Jl > b JS- (fori=l,2,3,...) 

9xj 3xj 3xj 

(2) ^SUiLg + f^L (fori- 1,2,8,...) 

9xj 3xj 9xj 

(3) 3(h(f)) = h , (f? 3f (fori= 1,2,3,...) 

dxj 3xj 

Equation (3) may be more recognizable if the composite functions are written as 
h o f and h' o f. AH these results follow from the similar rules for ordinary differen- 
tiation. Lists of useful derivatives are to be found in 

M231 Handbook, page 45 et seq. 
MST 282 Handbook, page 27. 

Example Find ^x + xz sin(x 2 y)) . 

3x 
We have 

3(2x + xzsin(x 2 y)) = 2 3x + 3(xz sin(x 2 y)) by result (1) 

3x 3x 3x 

-2 + W sin(x 2 y) + xz 3 < sin < X 'y )) . by result (2), 

3x 3x 

= 2 + zsin(x 2 y) + xz sin'(x 2 y) liLZ/, by result (3) 

3x 

= 2 + zsin(x 2 y) + 2x 2 yz cos(x 2 y). 



M334I.1 



Summary 

Notation 

g(f) 

go f 

f- 1 

E 3 
P 

(Pl>P2,P3) 
p + q 

(Pi»P2,P3) + (qi,q2,q3) 

ap 

a (Pl>P2>P3) 

x, y, z or x 1? x 2 , x 3 

af 3f , 3f 
_ , and 

dx dy dz 

f+g 

E 2 

E 1 =R 

Definitions 

(i) Euclidean 3-space E 3 

(ii) Sum of points of E 3 

(iii) Scalar product of a point of E 3 

(iv) Euclidean plane E 2 

(v) Real line E 1 = R 

(vi) Natural coordinate functions x, y, z or x t , x 2 , x 3 

(vii) Pointwise addition and multiplication of functions 

(viii) Composite of functions g(f) 

(ix) Differentiable function 

(x) Inverse function f -1 

Results 

Partial differentiation satisfies the following rules. 



(1) 
(2) 

(3) 



3(af+bg) = q 3f_ + b 3g_ 



- a 



bx[ 



3xj 9xj 



a(fg) = 9f f 

3xj 9xj 



+ f 



ag 



9xj 



a(h(0) = hV) 5L 

bx[ 9xj 



where a, b are real numbers, f, g : E' 



(i= 1,2,3,...) 
(i=l,2,3,...) 

(i= 1,2,3,...) 
->Randh:R ► R. 



Page 1 , line - 4 
Text, page 5 
Page 2, line 16 
Page 3, Definition 1.1 
Page 3, Definition 1.1 
Page 3, Definition 1.1 
Page 3, line -3 
Page 3, line - 3 
Page 3, line - 1 
Page 3, line -1 
Page 4, Definition 1.2 

Page 4, line 24 

Page 4, line -5 
Page 4, line - 5 
Page 5, line 22 
Page 5, line 23 



Page 3, Definition 1.1 
Page 3, line -5 
Page 3, line - 2 
Page 5, line 22 
Page 5, line 23 
Page 4, Definition 1.2 
Page 4, line - 5 
Page 1 , line - 4 
Page 4, Definition 1.3 
Page 2, line 16 



Text: page 5 



Techniques 

(i) Evaluation of a function. 

(ii) Simplification of a function. 

(iii) Partial differentiation, using results (1), (2), (3). 



M334I.1 
Exercises 

Technique (i) 

1. Page 5, Exercises 2(b) and 2(c). 

Technique (it) 

2. Page 5, Exercise 1(a). 

3. Page 6, Exercise 4(a). Just simplify this function. 

Technique (Hi) 

4. Page 5, Exercises 1(c) and 1(d). 

5. Page 6, Exercises 3(a) and 3(b). 

6. Page 6, Exercise 4(a). 



Solutions 

1. Page 5, Exercises 2(b) and 2(c). 

If p is a point of E 3 , then by the pointwise definition of addition and 
multiplication of functions 

f(p) = ((*(p)) 2 y(p» - ((y(p)) 2 *(p)). 

By Definition 1.2 

x((3,-l,i)) = 3, y((S,-l,i))=-l and z((3, -1, *)) =i 
Hence 

f((3,-l,i)) = (3 2 .(-l))-(("l) 2 -T)=-9i. 
As in the above result 

f((a, 1, 1-a)) = a 2 .l - l 2 .(l-a) = a 2 + a - 1, for any real number a. 

2. Page 5, Exercise 1(a). 

If f = x 2 y then, since f is defined by pointwise multiplication, we have 

f(p) = x 2 (p) y(p) = p\ p 2 , where p = (p u p 2 , p 3 ) is a point of E 3 . 

The function g = y sin z is defined in terms of composition and pointwise 
multiplication and so 

g(p) = y(p) sin ( Z (P)) = P2 sin p 3 . 
Hence 

(fg 2 )(p) = f(p)(g(p)) 2 

= P?P2 (P2 sinp 3 ) 2 
-P!P 2 sin 2 p 3 



M334I.1 
and 

fg 2 = x 2 y 3 sin 2 z. 

In this way we see that pointwise addition and multiplication behave exactly 

as expected. 

If f = x 2 y and g = y sin z then 

fg 2 = x 2 y(y sin z) 2 = x 2 y 3 sin 2 z. 



3. Page 6, Exercise 4(a). 

If h = x 2 - yz then, by the pointwise definitions of addition and multiplica- 
tion of functions 

h (p) = Pi " P2P3> for any point p = (p 1? p 2 , p 3 ) in E 3 . 

Since f(p) = h(g! (p), g 2 (p), g 3 (p)) it follows that 

f(p) = (gi(p)) 2 -(g 2 (p))(g3(p)). 

Hence f = g? - g 2 g 3 . 

In this exercise gi = x + y, g 2 = y 2 , g 3 = x + z and so 
f=(x + y) 2 - y 2 (x + z) 
= x 2 + 2xy + y 2 - xy 2 - y 2 z. 

4. Page 5, Exercises 1(c) and 1(d). 
Now fg = x 2 y 2 sin z. 

Treating x and y as constants we find that 

3(fg) 2 2 

_l_2i = x z y z cos z. 

3z 
Treating x and z as constants we find that 

a 2 (fg) 



9y9z 



= 2x y cos z. 



Now sin f = sin x 2 y. 

Treating x as a constant we find that 

9(sin f) 2 2 

— i 4_ = x cos x y. 

by 



Page 6, Exercises 3(a) and 3(b). 

If f = x sin (xy) + y cos (xz) then, treating y and z as constants, 

7\f 

= xy cos (xy) - yz sin (xz). 

9x 



M334I.1 9 

To deal with Exercise 3(b) we need to extend result (3) to cover the partial 
differentiation of the composite of three functions, h from E 3 to R and 
hi, h 2 from R to itself. In this exercise it will be more convenient to use the 
symbol o for the composite of functions. 

3 < h * < h ' oh » = (hi . (h, . h)) 8 < h ' ° h > . by result (3), 

3x 3x 

= (hi o (h t o h)). {h\ o h) j^l, by result (3). 

3x 
Here h = x 2 + y 2 + z 2 , hj = exp, h 2 = sin 

and f = sin o (exp o ( x 2 + y 2 + z 2 )) = h 2 o (hi o h). Hence 

11= (sin' o (exp o (x 2 + y 2 + z 2 )). exp' o (x 2 + y 2 + z 2 ). 8 (* 2 + Y* + z *) 
dx 3x 

, (x 2 +y 2 +z 2 ), (x 2 +y 2 +z 2 ) 
= cos(e v } ') e v 7 '2x 

_ g (x 2 +y 2 +z 2 ) , (x 2 +y 2 +z 2 ), 
= 2x e v } ' cos (e v 7 '). 

6. Page 6, Exercise 4(a). 

From Exercise 3 

f = x 2 + 2xy + y 2 - xy 2 - y 2 z 

hence 

— = 2x + 2y-y 2 . 
dx 



10 M334 1 .2 

1.2 TANGENT VECTORS 

Introduction 

In this section we review arrows. They will be called tangent vectors. At each point 
of Euclidean space they form a vector space. We also reviewvector fields. They are 
functions assigning to each point of Euclidean space a tangent vector based at that 
point. We generalize the notion of basis and coordinates to deal with vector fields. 

READ: Section 1.2 (pages 6-10). 

Comments 

(i) Page 9: Definition 2.4 Whereas a vector field V assigns one tangent 

vector to each point of Euclidean space, the natural frame field assigns three. 

(ii) Page 10: end of Section Since any vector field V can be expressed as 

V = SvjUj, for suitable real valued functions v l5 v 2 and v 3 the natural frame 
field does serve as a 'basis'. When we used the term basis for a real vector 
space it implied that every vector could be expressed as a linear combination 
of the basis vectors. The values of the vector fields Ui, U 2 , U 3 are linearly 
independent at each point. However, not every vector field can be expressed 
as a linear combination of the Uj's with constant coefficients. That is the 
coefficients in the sum Ev^U^ are not necessarily real numbers. Now they are 
real valued functions and the collection of real valued functions is not even a 
field since it is not possible to find a multiplicative inverse of a function 
which is zero at some points even if it is not always zero. These functions 
form a ring and the vector fields are a finite-dimensional module over this 
ring. 

Supplementary Comment 

(i) Page 10: lines 3 and 4 These are generalizations of the usual rules for the 

addition and scalar multiplication of vectors expressed in terms of co- 
ordinates. 
The first equation is obtained as follows: 

(2(v i + w i )U i )(p) = 2(vi + wi)(p)U i (p), by repeated use of the 

pointwise formulas for 
addition and scalar multi- 
plication of vector fields, 

= 2(vj(p) + wi(p))Ui(p), by the definition of point- 

wise addition of two 
functions, 

= 2vj(p)Ui(p) + 2wi(p)Ui(p), by the usual rules for a 

vector space, 
= (SviUiXp) + (2wiUi)(p) 

= (SvjUj + Zw}Uj)(p), by the pointwise rules for 

the "arithmetic" of 
vector fields. 

Since vector fields are defined as functions on E 3 , whenever two vector 
fields agree at each point they must be the same vector field. 



M334 1 .2 
Hence 

2(vi + wi)Ui = SvjUi + wiUj. 
The result 

2(fv i )U i = f(2:v i U i ) 
follows similarly. 



11 



Summary 

Notation 

V P 
T p (E 3 ) 

Vp + Wp 

av p 
V 

V + W 

tv 

{U!,U 2 ,U 3 } 
SviUi 



Page 6, Definition 2.1 

Page 7, Definition 2.2 

Page 7, line 9 

Page 7, line 9 

Page 8, Definition 2.3 

Page 8, line -8 

Page 8, line -3 

Page 9, Definition 2.4 

Page 9, line -8 



Definitions 



i) 

") 
iii) 

iv) 

v) 
vi) 

vii) 



Tangent vector v p 

Vector part of tangent vector 

Point of application of tangent vector 

Parallel tangent vectors 

Tangent space T p (E 3 ) 

Addition and scalar multiplication of tangent vectors 

Vector field V 
viii) Pointwise principle 
ix) Addition of vector fields 



x) Multiplication of a vector field by a real-valued 
function 

xi) Natural frame field {U^ U 2 , U 3 } 

xii) Euclidean coordinate functions 

xiii) Differentiable vector field 



Page 6, 


Definition 2.1 


Page 6, 


Definition 2.1 


Page 6, 


Definition 2.1 


Page 6, 


line -5 


Page 7, 


Definition 2.2 


Page 7, 


line 9 


Page 8, 


Definition 2.3 


Page 8, 


line -7 


Page 8, 


line -10 


Page 8, 


line -3 


Page 9, 


Definition 2.4 


Page 9, 


Lemma 2.5 


Page 10, line 6 



12 M334 1.2 

Results 

(i) Tp(E 3 ) is linearly isomorphic to E 3 . Page 7, line -3 

(ii) Each vector field V can be uniquely expressed as 

V = SvjUj, where v l5 v 2 , v 3 are real-valued functions. Page 9, Lemma 2.5 

(iii) Each tangent vector (a 1} a 2 , a 3 ) p can be expressed 

in terms of the natural frame field as 2ajUi(p). Page 9, line -5 

(iv) Addition and multiplication by a function are 
expressed in terms of coordinates by 

SviUj + ZwiUi = 2(vj + wi)Uj 

f(SviUi) = S(fvi)Ui- Page 10, lines 3 and 4 



Techniques 

(i) Representation of tangent vectors in standard form 

Vp. Page 6, Definition 2.1 

(ii) Addition and scalar multiplication of tangent 

vectors. Page 7, line 9 

(iii) Evaluation of a vector field at a point. Page 8, Definition 2.3 

(iv) Addition and multiplication, by real-valued 

functions, of vector fields. Page 8, line -8 and 

Page 9, line 2 

(v) Pictorial representation of tangent vectors and 

vector fields. Page 7, Fig. 1.1 

(vi) Representation of tangent vectors and vector fields 
in terms of the natural frame field and Euclidean 
coordinates. Page 9, Lemma 2.5 

(vii) Addition and multiplication, by a real-valued 
function, of vector fields expressed in terms of 
Euclidean coordinates. Page 10, lines 3 and 4 



Exercises 

Techniques (ii), (vi) and (v) 

1. Page 10, Exercise 1. 

Techniques (iii) and (vii) 

2. Page 10, Exercise 2. 

Techniques (vii), (vi), (iv) and (i) 

3. Page 10, Exercise 3. 



M334 1.2 
Solutions 

1. Page 10, Exercise 1. 

(a) 3v- 2w = 3(-2, 1,-1)- 2(0,1,3) 

= (-6,3,-3)- (0,2,6) 
= (-6,1,-9). 

By the definition of addition and scalar multiplication of tangent vectors, 

3v p - 2w p = (3v - 2w) p = (-6, 1, -9) p . 
Using the identity (a 2 , a 2 , a 3 ) p = SajU^p), we find that 

3v p - 2w p = -eU^p) + U 2 (p) - 9U 3 (p). 

(b) Vp represents the arrow from the point p to the point p + v. 
Herep = (l, l,0),v = (-2, 1, -1) and hence p + v = (-1, 2, -1). 
Similarly, w p is the arrow from p = (1, 1, 0) to p + w = (1, 2, 3). 

Now 



13 



-2v p = (-2v)p = (4,-2,2)p 



and 



v p + w p = (v + w) p = (-2, 2, 2) p 
and hence they are the arrows from (1, 1, 0) to (5, -1, 2) and (-1,3, 2). 



^^^(-1,3,2) 



(5,-1,2) 




14 M334 1.2 

2. Page 10, Exercise 2. 

W - xV = 2x 2 U 2 - U 3 - x(xUi + yU 2 ) 
= 2x 2 U 2 - U 3 - x 2 U!- xyU 2 
= -x 2 U 1 + (2x 2 -xy)U 2 -U 3 , 

by result (iv) on the addition and multiplication of vector fields in terms of 
coordinates. 

If p = (-1, 0, 2) then x(p) = -1, y(p) = and z(p) = 2. 

By the definition of the natural frame field 

U^p) = (1,0, 0)p, U 2 (p) = (0, 1, 0) p , U 3 (p) = (0, 0, l) p . Hence by the pointwise 
definition of adaition and multiplication 

(W - xV)((-l, 0, 2)) = -(-1) 2 (1, 0, 0) p + (2.(-l) 2 - (-1).0)(0, 1, 0) p - (0, 0, l) p 

= (-l,2,-l) p . 

3. Page 10, Exercise 3. 

(a) Rearranging and dividing by 7 we obtain 

7 7 

(b) V(p) = ( Pl ,p 3 -p 1 ,0) p 

= (xj(p), x 3 (p) - x^p), 0) p , using the definition of the 

coordinate functions, 

= (x^p), (x 3 - X!)(p), 0) p , by the definition of linear 

combinations of func- 
tions, 

= x 1 (p)U 1 (p) + (x 3 - Xi)(p)U 2 (p), by the standard identity 

(a 1 ,a 2 ,a 3 ) p = 2:aiUi(p). 

Hence, using the definition of linear combinations of vector fields, 

V = x 1 U 1 + (x 3 -x 1 )U 2 . 
Alternatively we could write this as 

V = xU! + (z- x)U 2 . 

(c) V=2(xU 1 + yU 2 )-x(U 1 -y 2 U 3 ) 

= 2xUj + 2yU 2 - xUi + xy 2 U 3 , by result (iii) on addition 

and multiplication in 
terms of coordinates, 

= (2x - x)U! + 2yU 2 + xy 2 U 3 

= xU 1 + 2yU 2 + xy 2 U 3 . 

(d) If p = (pi, p 2 , p 3 ) then V(p) is the vector from (p t , p 2 , p 3 ) 
to(l+ Pl , p 2 p 3 , p 2 ). 



M334 1.2 15 

Hence V(p) = ((1 + Pl , p 2 p 3 , p 2 ) - (p„ p 2 , p 3 )) p 

= ( 1 »P2P3~ P2»P2~ P3)p- 

By arguments similar to those in part (b) we find that 

V = Uj + x 2 (x 3 - 1)U 2 + (x 2 - x 3 )U 3 
or alternatively 

V = U 1 + y(z-l)U 2 +(y-z)U 3 . 

(e) The vector from p to the origin is - p = -p so 
V (P) = (-P)p = H>i> "P2, -P 3 )p- 

Hence V = - x^ - x 2 U 2 - x 3 U 3 or alternatively 
= - xU! - yU 2 - zU 3 . 



16 M334 1.3 

1.3 DIRECTIONAL DERIVATIVES 

Introduction 

This section follows on from Sections 1.1 and 1.2. In Section 1.1 we reviewed partial 
differentiation. We recalled that for any differentiable function f on E 3 and for any 
point p in E 3 

f^p)=-|:(f(t,P*P3))lt = p 1 - 

Instead of evaluating this derivative at t = p x we often 'shift' the function so that we 
can evaluate the derivative at t = 0. That is 

— (p) = — (%i + t, p 2 , p 3 )) It = 0- 
9xj dt 

Using the natural frame field introduced in Section 1.2 we can write this as 

~(p) =^(f(P + tu 1 ( P )))| t = . 

There are similar formulas for (3f/3x 2 )(p) and (of/3x 3 )(p), so to find (8f/3xj)(p) we 
find the initial rate of change of the function f in the direction Ui(p). In Unit 
MST282 7, Work and Energy I, Section 7.1.2 we met a generalization of this 
technique. If n was a unit direction vector we defined the directional derivative, 
(3f/3n)(p), to be the initial rate of change of f in the direction n and proved that 

-- = n-Vf(p) = Sni — (p). 
on i=i 3xj 

In this section we generalize the above definition and result to deal with any 
tangent vector v p and then any vector field V. We show that these directional 
derivatives satisfy properties similar to those of ordinary derivatives. 

READ: Section 1.3 (pages 1 1-14) omitting the proof of Lem ma 3.2. 

Comments 

(i) Page 12: Lemma 3.2 This is a generalization of the result mentioned in 

the introduction. The directional derivative is 

v p [f]=Wf( P ) = l;v i i£(p). 

i=i oxj 
It is proved in exactly the same way using a version of the 'chain rule'. This 
will be dealt with in Section 1.7 and until then you should accept this result 
without proof. 

(ii) Page 12: Theorem 3.3.(3) The symbol * in this theorem and in Corollary 
3.4 signifies the ordinary product of two numbers. The inner product of two 
vectors will not be introduced until Chapter II. 

(iii) Page 12: line 15 to line -3 The real-valued function V[f] is defined by 

(V[f])(p)=V(p)[f]. 



M334 1.3 17 

This equation is used in the proof of Corollary 3.4. 

For computational purposes the identities Uj[f] = 3f/3xj (i = 1, 2, 3) will 
be very important. We can obtain these results from Lemma 3.2. For 
instance 



3f 



3f 



3f 



3f 



Ui(p) [f] =U,0,0) D [f] =l.^L(p) + 0. (p) + 0. (p)- (p). 



3xj 



9xo 



dx-. 



dxj 



(iv) Page 14; line 14 to line 21 This is a very important remark. For vector 

fields V and W we consider combinations of the form fV + gW, where f and g 
are real-valued functions. For functions f and g we consider combinations of 
the form af + bg, where a and b are real numbers. 

In the example 

(xUi -y 2 U 3 )[x 2 y + z 3 ] = xU 1 [x 2 y + z 3 ] - y 2 U 3 [x 2 y + z 3 ] , 

by Corollary 3.4.(1), 

= xU 1 [x 2 y] +XUJZ 3 ] -y 2 U 3 [x 2 y] -y 2 U 3 [z 3 ], by Corollary 3.4.(2). 

Supplementary Comment 

(i) Page 12: Theorem 3.3.(1) and (2) 

(1) If v = (vj, v 2 , v 3 ) and w = (w t , w 2 , w 3 ) then 

(avp + bwp)[f] = (av + bw)p[f], by the definition of addition 

and scalar multiplication of 
tangent vectors, 

= 2(avj + bwj) (p), by Lemma 1.3.2, 

9xj 

= a(2v i ^L(p)Ub(2w i ^L(p) 



dxj 



3xj 



(2) 



= av p [f] +bw p [f]. 
v p [af + bg] =S Vi | 3(af + b gM (p), by Lemma 1.3.2, 

^vifa^pj.b^p)) 

= af2v i £.(p)j + b/2v i iL(p)|. 



Summary 



Notation 

v p [f] 
V[f] 



Page 11, Definition 3.1 
Page 13, line 15 



18 M334 1.3 

Definitions 

(i) Derivative of a function with respect to a tangent 

vector v p [f] . Page 11, Definition 3.1 

(ii) Derivative of a function with respect to a vector 

field V[f]. Page 13, line 15 

Results 

(i) If v p = (vj, v 2 , v 3 ) p then v p [f] = 2vj— (p). Page 12, Lemma 3.2 

(ii) Directional derivatives, with respect to tangent 
vectors and vector fields, satisfy linear and 
Leibnizian properties: 

(a) (av p + bw p ) [f] = av p [f ] + bw p [f ] . 

(b) v p [af + bg] =av p [f] +bv p [g]. 

( c ) v p [fg] = v p [f] g(p) + f(p)v p [g] . Page 12, Theorem 3.3 

(a) (fV + gW)[h] = fV[h] + gW[h] . 

(b) V[af + bg] = aV[f] + bV[g] . 

(c) V[fg] = V[f] . g + fV[g] . Page 13, Corollary 3.4. 

(iii) Directional derivatives with respect to components 
of the natural frame field are partial derivatives 

Uj[f] =— (i = 1, 2, 3). Page 13, line -10 

Techniques 

Evaluation of directional derivatives with respect to 
tangent vectors or vector fields using: 

(i) the definition, for tangent vectors Page 11, Definition 3.1 

(ii) the definition, for vector fields Text, page 16 

(iii) coordinates and partial derivatives Page 12, Lemma 3.2 

(iv) the linear and Leibnizian properties and the 

relationship between the natural frame fields and 

partial derivatives. Page 12, Theorem 3.3, 

Page 13, Corollary 3.4 and 
Page 13, line -10. 

Exercises 

Technique (i) 

1. Page 14, Exercise 1(c). 

Technique (ii) 

2. Determine V[f] where f = e x cos y and V = 21^ - U 2 + 3U 3 . 



M334 1.3 19 

Technique (Hi) 

3. Page 15, Exercise 2(c). 

Technique (iv) 

4. Page 15, Exercise 3(e). First prove that Vtf 2 ] = 2f V[f] . 

Theory Exercises 

5. Page 15, Exercise 4. 

6. Page 15, Exercise 5. (HINT: Use Exercise 4.) 



Solutions 



Page 14, Exercise 1(c). 
By Definition 1.3.1, 

vp[f]=~(f(p + tv))| t = 0. 

Here 

p + tv = (2, 0, -1) + t(2, -1, 3) = (2 + 2t, -t, -1 + 3t) 
and 

f(p + tv) = (e x cosy) (2 + 2t, -t, -1 + 3t) 

= e 2 + 2t cos (-t), by the pointwise definition of addi- 
tion, multiplication and composition 
of functions and the definition of the 
functions x and y, 
= e 2 e 2t cos t. 
Hence 

d 



v p [f]=e 2 ^(e 2 W)| t = 

= e 2 (2e 2t cos t - e 2t sin t)| t _ q = 2e 2 . 
The value of the function V[f] at a point p is given by the formula 

(v[f]Xp) = v(p)[f]. 

Now V(p) = 2U,(p) - U 2 (p) + 3U 3 ( P ) = (2, -1, 3) p . 
Hence if p = (p t , p 2 , p 3 ) then 

V( P )[f] =4(f((Pi, P 2 , Pa) + t(2, -1, 3)))| t = 
dt 

= (f( Pl + 2t,p 2 -t,p 3 +3t))| t = o 
dt 

_ d, p! + 2t . . Xl 

~-r( e cos (P2 " *)) t = 

at ' 

= (2ePi + 2t cos(p2 . t) + ePl + 2t sin(p2 _ t )) 1 1 = 
= e" 1 (2cos p 2 + sin p 2 ). 



20 M334 1.3 

So V[f] (p) = eP»(2cos p 2 + sin p 2 ) and hence 
V[f] = e x (2cos y + sin y). 

3. Page 15, Exercise 2(c). 
By Lemma 1.3.2 

v p [f]=Sv i JL( P ), 
3xj 

though in this solution we replace x lt x 2 , x 3 by x, y, z. Since f = e x cos y it 
follows that 

af - x 9 f x- a 9f n 
e A cos y, = -e x sm y and = 0. 

3x dy 9z 

Since p = (2, 0,-1) 

—(p) = e\ il(p) = and il(p) = 0. 
9x 8y 9z 

Hence, since v = (2, -1, 3) 

v p [f] = 2.e 2 + (-l).O + 3.0 = 2e 2 . 

4. Page 15, Exercise 3(e). 

By the Leibnizian property, Corollary 1.3.4(3), 

V[f 2 ] =V[f.f] =V[f].f + fV[f] =2fV[f]. 
Hence by the linearity property, Corollary 1.3.4(2), 

V[f 2 + g 2 ]=2(fV[f]+gV[g]). 
Now 

V[f] =(y 2 U,-xU,)[xy] =y 2 U,[xy] -xU 3 [xy] 

= y^lM- x ^I), since UilH =— , 
9x 9z 9xj 

= y 2 .y - x.O = y 3 . 
Similarly, 

V[g]=(y 2 U 1 -xU 3 )[z 3 ] 
= -3xz 2 
and hence 

V[f 2 + g 2 ] =2(xy.(y 3 ) + z 3 .(-3xz 2 )) 
= 2x(y 4 - 3z 5 ). 

5. Page 15, Exercise 4. 

Following the hint we evaluate V[x;] , where V = ZvjUj. 

For fixed j we have 

V[x j ]=2v i U i [x j ]=2v i ^. 

9xj 



M334 1.3 21 

dx- 
Now J = 1 if i = j and is zero otherwise. Hence 

dxj 

V[xj] =vj for all j, 

and 

V = 2v i U i = SV[x i ]U i . 

6. Page 15, Exercise 5. 

Since V[f] = W[f] for every function f on E 3 it follows that 

V[ Xi ]=W[xi] (i = l,2,3). 

Hence by the result of Exercise 4 

V = SV[xi] Ui = 2W[ Xi ] Ui = W. 



22 



M334 1.4 



1.4 CURVES IN E 3 



Introduction 



This section follows on from Sections 1.2 and 1.3. We review two ways of defining 
curves, the definition of the velocity of a curve, and show how the rate of change of 
a function along a curve is related to the directional derivative with respect to the 
velocity vector. 

READ: Section I. 4 (pages 15-21) omitting the proof of Lemma 4.6. 
Comments 



(i) Section 1.4 From Definition 4.1 you might think that curves always 

have routes that appear in some sense smooth. However, curves can be 
degenerate in several different ways. For instance, the simplest curve is the 

constant curve a : 1 1 ► 0, which has as its route a single point. The curve 

01 : t ' — > (sin t, 0, 0) has as its route a closed interval which it traverses 
infinitely often. It is even possible to have a curve that turns sharp corners. 
For instance, the curve 



'(-e-Vt'.o.O) 

a : 1 1 — ► \ (0, 0, 0) 

has the following route 



for t < 
f or t = 
for t > 



V(0,1,0) 



(-1,0,0) 



(0, 0, 0) 



>x 



This function was chosen to ensure that the Euclidean coordinate functions 
are 'infinitely' differentiable. In order for the curve to stand still, reverse or 
turn a sharp corner, as in the above examples, it is necessary for the velocity 
vector to take the value zero. Hence, if we restrict attention to regular curves 
all these pathologies can be avoided. The definition of a 'Curve' ensures that 
we can always find a regular parametrization of each component. We use the 
word "closed" to describe components of a Curve that are like a distorted 
circle and so not surprisingly they have periodic parametrizations. Any 
component that is not like a distorted circle is like a distorted copy of the 
real line and so not surprisingly has a one-to-one parametrization. The circle 
and a branch of the hyperbola, described on page 21, are typical of closed 
and not closed Curves. 



M334 1.4 



23 



Page 15: line 9 of the section Here the Euclidean coordinate functions 
are real-valued functions on some interval I. When we described a vector field 
in terms of its Euclidean coordinate functions they were real-valued func- 
tions on E 3 . 



Supplementary Comments 

(i) Page 16: Example 4.2(3) 

a(t) = (2cos 2 t, sin 2t, 2sin t) = (2cos 2 1, 2sin t cos t, 2sin t). 
Since 

(2cos 2 t) 2 + (2sin t cos t) 2 + (2sin t) 2 

= 4cos 2 t(cos 2 t + sin 2 1) + 4sin 2 t = 4(cos 2 t + sin 2 t) = 4 

it follows that a(t) belongs to S. 

Also 

a(t) - (1,0,0) = (2cos 2 t - 1, sin 2t, 2sin t) = (cos 2t, sin 2t, 2sin t) and since 
cos 2 2t + sin 2 2t = 1 it follows that a(t) belongs to C. 

As t tends to the point a(t) tends to (2,0,0), and as t tends to tt/2 the point 
a(t) tends to (0,0,2), and hence ex. follows the route sliced from C by the 
sphere S. 

(ii) Page 18: line 6 Since cq(t) = pj + qj (i = 1, 2, 3) it follows that 

dai/dt(t) = qj (i = 1, 2, 3) and hence oc'(t) = (q lf q 2 , q 3 )a(t)- 

(iii) Page 19: Lemma 4.5 By the definition of the composite of two 

functions, 

(<*(h))(s) = a(h(s)) and («i(h))(s) = <*i(h(s)). 

Hence, since j3(s) = a(h(s)) = (a(h))(s), 

0'(s) = (a(h))'(s). 

Since a(h(s)) = (a 1 (h(s)), a 2 (h(s)), a 3 (h(s))), 

(«(h))(s) = ((a,(h))(s), (a 2 (h))(s), (a 3 (h))(s)) 

and by Definition 4.3 

(a(h))'(s) = ((«i(h)) f (s), (a 2 (h))'(s), (a 3 (h))'(s)). 

The result now follows as on page 19. 



Summary 

Notation 

<*(*) = Wt),a 2 (t),a 3 (t)) 
a = (<*i, <*i, « 3 ) 

OL 
«'(t) 

C: f=a 



Page 15, line 7 
Page 15, line 10 
Page 15, Definition 4.1 
Page 17, Definition 4.3 
Page 20, line -5. 



24 
Definitions 

(i) Euclidean coordinate functions of a curve 

(ii) Curve a. 

(iii) Velocity vector of a curve a'(t) 

(iv) Reparametrization of a curve 

(v) Periodic curve 

(vi) Period of a curve 

(vii) Regular curve 

(viii) 'Curves' on the plane defined implicitly C: f = a 

(ix) Closed curve on the plane 



M334 1.4 



Page 15, line 9 

Page 15, Definition 4.1 

Page 17, Definition 4.3 

Page 18, Definition 4.4 

Page 20, line 13 

Page 20, line 15 

Page 20, line -10 

Page 20, line -5 

Page 21, line 5 and 
Text: page 22 



Examples 

(i) Straight line. a(t) = p + tq 

(ii) Helix. a(t) = (a cos t, a sin t, bt) 



Page 15, Example 4.2.(1) 
Page 15, Example 4.2.(1) 



Results 



(i) If ]8 is the reparametrization of a by h, then 

^(s)=ffl(s).«'(h(s)). 
Ids/ 

(ii) If ex is a curve in E 3 and f is a differentiable 
function on E 3 , then 

„. (t)[f] =M (t) 



Page 19, Lemma 4.5 



Page 19, Lemma 4.6 



Techniques 

(i) Determination of a parametric representation of a 

straight line. 

(ii) Determination of the velocity vectors of a curve. 

(iii) Determination of the effect of reparametrization 
on velocity vectors. 

(iv) Calculation of the rate of change of a function 
along a curve by means of the directional deriva- 
tive with respect to the velocity vector. 

(v) Parametrization of implicitly defined 'Curves' in 

E 2 . 



Page 15, Example 4.2.(1) 
Page 17, Definition 4.3 

Page 19, Lemma 4.5 

Page 19, Lemma 4.6 
Page 21. 



M334 1.4 25 

Exercises 

Technique (i) 

1. Page 21, Exercise 5. 

Technique (ii) 

2. Page 21, Exercise 1. 

Technique (Hi) 

3. Page 21, Exercise 3. Evaluate <*'(7r/4), 0'(l/v/2) and (dh/ds)(l/ v /2). Verify 

Lemma 1.4.5 in this case. 

l 

(Note: sin" 1 was called arc sin in M231 and (sin" 1 )' (s) = 1/(1 - s 2 ) 2 ".) 

Technique (iv) 

4. Page 21, Exercise 7. 

Technique (v) 

5. Page 22, Exercise 10. 

Solutions 

1. Page 21, Exercise 5. 

Suppose the line is given by a(t) = p + tq. If it passes through the points Pi 
and p 2 we can assume a(0) = pj and a(l) = p 2 . Then p = Pi and q = p 2 - pi 
and hence a(t) = Pi + t(p 2 - Pi). If Pi = (1,-3, -1) and p 2 = (6, 2, 1) the curve 
is 

a(t) = (1, -3, -1) + t((6, 2, 1) - (1, -3, -1)) 

= (l,-3,-l) + t(5,5,2) 

= (l + 5t, -3 + 5t,-l+2t). 

Similarly the line which passes through (-1, 1,0) and (-5, -1,-1) is given by 

0(s) = (-1 - 4s, 1 - 2s, -s). 

If the lines meet, there are real numbers t and s such that 

«(t)=/3(s). 

i.e. 1 + 5t = -1 - 4s 

-3 + 5t = 1 - 2s 

-1 + 2t = -s. 

The first two equations can be solved uniquely giving t = 2, s = -3, which is 
also a solution of the third equation, so the lines meet at 
a(2)=0(-3) = (ll,7,3). 



26 M334 1.4 

2. Page 21, Exercise 1. 

a(t) = (2cos 2 t, sin2t, 2sint). 

Hence by Definition 1.4.3, 

a'(t) = (£(2cos 2 1), _£_(sin 2t), jL(2sin t)\ 

Ut dt dt Mt) 

= (-4cos t sin t, 2cos 2t, 2cos t) ( 2cc .s 2 1, sin 2t, 2sin t) 



and 



""Q' ("2. °V2) (1,1,^2)- 



3. Page 21, Exercise 3. 

By Definition 1.4.4 the reparametrized curve is given by 

|3(s) = a(h(s)) = a(sin _1 s). 

Now a(t) = (2cos 2 t, sin 2t, 2sin t) and to make progress we express a(t) as a 
function of sin t. 

We find that a(t) = (2(1 - sin 2 t), 2sin t (1 - sin 2 t)*,2sin t), using the identi- 
ties cos 2 t + sin 2 t = 1 and sin 2t = 2sin t cos t. 

Hence |8(s) = o^sin" 1 ^)) 

= (2(l-s 2 ),2s(l-s 2 )t2s). 

The function sin" 1 has domain [-1, 1] and so is well defined for < s < 1 
and takes values in the interval < s < 7r/2. 

Now 

a'(t) = (-4cos t sin t, 2cos 2t, 2cos t) 

and 

«'I1\=U.±.-L, 0, 2._L) = (-2,0,V2). 
U/ \ n/2^2 y/21 

Also 

^(s) = (-4s,2(l-s 2 f-_i!L 1 ,2) 
\ (1-s 2 ) 2 / 

and 

/3'[_L-1 = (-2V2, 0, 2). 



"P 



Finally 



%)--L^ and iH-lW- 
ds (1-s 2 ) 2 dsV2/ 



1 \ _ • -i / 1 \ it 



Hence 

j3'[ 1 = [ ] • <*'(— ] and, since h ( x ] = sin 

V2/ ds\V2/ U/ V2/ \V2/ 4 

this agrees with Lemma 1.4.5. 
4. Page 21, Exercise 7. 

Let a(t) = (t, 1 + t 2 , t), 
0(t) = (sint, cost, t), 
7(t) = (sinht, cosht, t). 



M334 1.4 ' 27 

Then a'(0) = 0'(O) = 7'(0) = (1, 0, 1) (q, 1, 0)- The curves have the same 
initial velocity v p . 

By Theorem 1.4.6 

v p [f] =-l(f(«))|t = =-W))\t = = -(f(7))|t = 
dt dt dt 

and since f = x 2 - y 2 + z 2 we have 

_£(f(a))| t=: o=i(t 2 -(l + t 2 ) 2 + t 2 )| t = o = i(-l-t 4 )| t = o=0, 
dt dt dt 

-(f(»)|t = = l(sin 2 t - cos 2 t + t 2 )| t = o=0, 
dt dt 

-l(f(T))|t = = -i(sinh 2 t - cosh 2 t + t 2 )| t = = 0. 
dt dt 

This is the answer we expect since by Lemma 1.3.2 

(i. o. i) (0, 1, 0)M = (i— + o.ii + i.-W 1, 0)) 

\ 3x 3y 3z/ 

= 2x((0,l,0)) = 0. 



5. Page 22, Exercise 10. 

(a) a : 1 1 — ► (•§ cos t, sin t) 




28 



M334 1.4 



(b) « : t h— t> 



l-3t 




(c) art i — ►(t, e 1 ) 




M334 1.4 



(d) 



a: ti — ► 
0<t<l 



(t, (1 - t*)*) 



29 



>x 



30 M334 1.5 

1.5 1 -FORMS 

Introduction 

This section follows on from Section 1.3, Directional Derivatives. 

In this section we introduce 1-forms. A 1-form is an extension of the idea of linear 
functional. A linear functional is a linear transformation between a vector space V 
and the underlying vector space of reals R. Now we have a collection of vector 
spaces, the tangent spaces T p (E 3 ), and we require our 1-forms to map any tangent 
vector to a real number in such a way that the restriction to any one tangent space is 
a linear functional. 

READ: Section I. 5 (pages 22-25). 
Comments 

(i) Page 23: Definition 5.2 If V is a vector field, then the definition of the 

differential implies that df(V) = V[f]. This arises from the pointwise 
definition in the following way. For any point p in E 3 

(df(V))( P ) = df(V( P )) = V(p)[f] = (V[f] )(p) 

and so df(V) =V[f] . 

(ii) Page 23: Example 5.3.(1) The symbol 5jj is a very useful tool and 

identities such as 

SvjSy = Vl 8 h + v 2 5 i2 + v 3 5 i3 = Vi (i = 1, 2, 3) 
J 

occur frequently. To check this identity we look at an example, say i = 2. 
Then 

2 vj5 2J - = Vl 5 21 + v 2 S 22 + v 3 5 23 
J 

= Vj .0 + v 2 .l + v 3 .0 = v 2 . 

(iii) Page 23: line -1 We use the same symbol for a real number and for the 

constant function taking that value. Hence we can write dxj (Uj) = fa. We 
obtain this result from first principles as follows 

dx i (U j ) = U j [x i ]=^i = 5 ij . 
3 Xj 

This is the basic result to remember. It can be used to obtain the results of 
both Example (1) and (2) as follows: 

(1) dxi(vp) = dxi/svjUjtp)! = 2v j dx i (U j (p)) 



= ZvjSij = Vi . 
J 



(2) 



*(vp) = fcfidxij (evjUjCp) 1 



2f i (p)v j dx i (U j (p))= Sf i (p)vj8 ij 

Zfi(p)vi. 
i 



09 



M334 1.5 



31 



(iv) Page 25: Lemma 5. 7 This follows from the rule for the partial differen- 
tiation of a composite given in the Additional Comments to Section 1.1. 



Supplementary Comments 

(i) Page 23: lines 5 to 10 Two functions are equal if they take the same 

value at each point p of E 3 . 

Now 

(0(fV + gW))(p) = 0((fV + gW)(p)), by the definition of the 

evaluation of a 1-form on a 
vector field, 

by the pointwise definition 
of operations on vector 
fields 



= 0(f(p)V(p)+g(p)W(p)), 



= f(p)0(V(p))+g(p)0(W(p)), 

= f(p)Wv))( P ) + g(p)(0(w))(p) 

= (f0(V) + g0(W))(p). 
Hence 

0(fV + gW) = f 0(V) + g 0(W). 
Similarly, since 

((f0 + g 0)(V))(p) = (f0 + g 0)(V(p)) 

= f(p)0(V(p))-f g (p)0(V(p)) 

= f(p)(«v))(p) + g(p)(WV))(p) 

= (f0(V) + g0(V))(p), 
it follows that (f0 + g0)(V) = f0(V) + g0(V). 



by the definition of a 
1-form 



(ii) Page 23: Definition 5.2 The function df is a 1-form since 

df(av p + bw p ) = (av p + bw p )[f] 

= av p [f] +bw p [f], by Theorem 1.3.3.(1), 

= adf(vp) + bdf(w p ). 



(iii) Page 24: the proof of Lemma 5.4 Since the natural frame field forms a 
basis, when restricted to each vector space, any 1-form that is zero on all 
three of these vector fields must be the zero 1-form. For j = 1, 2, 3, 



- 20(Ui)d Xi (Uj) = 0(Uj) - SflUiJdxifUj) 

= 0(Uj) - ZflUOSij = 0(Uj) - 0(Uj) = 0. 
Hence - 20(Ui)dxi = and = 20(U i )dx i . 



32 M334 1.5 

(iv) Page 24: the proof of Corollary 5.5 This can be proved directly using 
Lemma 1.5.4. 

df = 2df(Ui)dxi, by Lemma 1.5.4, 
i 

= 2Uj[f] dxi, by Definition 1.5.2, 

i 

= 2 dxj. 

idxj 

(v) Page 25: the worked example How do we obtain the first term 2xy dx? 
Now f = x 2 y + . . . and hence df = d(x 2 y) + • • • = <*(x 2 )y + x 2 dy + . . . , by 
Lemma 1.5.6. To evaluate d(x 2 ) rigorously we need to introduce the 
squaring function S : 1 1 — ► t 2 for which S' : 1 1 — ► 2t. 

Then d(x 2 ) = d(S(x)) = S'(x)dx, by Lemma 1.5.7, 

= 2x dx. 

Hence df = 2xy dx + x 2 dy + . . . 

Finally v p [f] = df(v p ) is evaluated using the result of Example 5.3.(2). 

Summary 



Notation 


«Vp) 

0p 

f0 

0(V) 
df 

df(v p ) 

df(V) 

dxj 



Page 22, Definition 5.1 
Page 22, Definition 5.1 
Page 22, line -11 
Page 22, line -7 
Page 22, line -4 
Page 22, line -2 
Page 23, Definition 5.2 
Page 23, Definition 5.2 

Text: page 30 

Page 23, Example 5.3.(1) 

Page 23, line -12 



Definitions 

(i) 1-form 

(ii) Addition of 1 -forms 

(iii) Multiplication of a 1-form by a real-valued 

function 
(iv) The effect of a 1-form on a vector field 
(v) The differential of a real valued function df 

(vi) Kronecker delta 5jj 



Page 22, Definition 5.2 
Page 22, line -7 

Page 22, line -4 
Page 22, line -2 
Page 22, Definition 5.2 
Page 23, line -12 



M334 1.5 33 

Results 

(i) 0(V) is linear in both the 1 -forms and vector 

fields V. Text: page 31 

(ii) dxi(vp) = vj. Page 23, Example 5.3.(1) 

(iii) If = Zfidxi, then 0(v p ) = S^p)^. Page 23, Example 5.3.(2) 

(iv) dxj(Uj) = 5jj. Page 23, line -1 

(v) = 20(Ui)dxj. Page 24, Lemma 5.4 

(vi) df = 2 dxj. Page 24, Corollary 5.5 

9xj 

(vii) d(f+g) = df+dg. Page 24, line -9 

(viii) d(fg) = gdf + fdg. Page 24, Lemma 5.6 

(ix) If f : E 3 ► R and h : R >R then d(h(f)) = h'(f)df. Page 25, Lemma 5.7 

Techniques 

(i) Recognition of a 1-form, by comparison with the 

formula 0(av p + bw p ) = a0(v p ) + b0(w p ). Page 22, Definition 5. 1 

(ii) Expressing a 1-form in standard form using the 

formula = 20(Ui)dxi. Page 24, Lemma 5.4 

(iii) Evaluation of the effect of a 1-form on a tangent 
vector using the formula 

0( v p) = 2fi(p)vi where = S^dx;. Page 23, Example 5.3.(2) 

i i 

(iv) Evaluation of the effect of a 1-form on a vector 

field using the identity dxi(Uj) = 5jj and linearity. Text: page 30 

(v) Calculation of differentials using the 'chain rule' 

At- 9f a Jf j 9f J 

df ~ dx i + dx 2 + dx 3 . Page 24, Corollary 5.5 

3xj dx 2 3x 3 

(vi) Calculation of differentials using the linear, 
Leibnizian and composite properties 

d(f + g) = df + dg, Page 24, line -9 

d ( f g) = g df + f dg, Page 24, Lemma 5.6 

d(h(f)) = h'(f)df. Page 25, Lemma 5.7 

(vii) Evaluation of directional derivatives using 
differentials 

v p[ f ] = d f(v p ). Page 23, Definition 5.2 



Exercises 



Techniques (i) and (ii) 

1. Page 26, Exercise 7. 



34 M334 1.5 

Technique (Hi) l 

2. Page 25, Exercise 1(c). 

Technique (iv) 

3. Page 25, Exercise 3. Evaluate on W only. 

Technique (v) 

4. Page 26, Exercise 5(a). 

Technique (vi) 

5. Page 26, Exercise 5(b). [Note: (tan" 1 )' (t) = 



-ix'- 1 



\ 1+V 

Technique (mi) 

6. Page 26, Exercise 6(b). 



Solutions 



Page 26, Exercise 7. 

To be a 1-form, needs to be linear on each tangent space Tp(E 3 ). That is, 
the mapping 

(vi, v 2 , v 3 ) i — ► ((vi, v 2 , v 3 )p) must be linear for each p in E 3 . 

(a) The mapping is 

(Vi, V 2 , V 3 ) I ► Vj - v 3 , 

which is a linear mapping from E 3 to R independent of whichever 
point p we choose. Hence this does give us a 1-form. 

Now0(U 1 (p)) = 0((l,O,O) p ) = l 

0(U 2 (p)) = 0((O,l,O)p) = O 

0(U 3 (p)) = 0((O, 0, l) p ) = -1 

and hence, since = 20(Uj)dxj, 
i 

= l.dx! - l.dx 3 = dxj - dx 3 . 

(b) This is not linear, since, for a fixed point p withp^ps and 
non-zero vector v, the definition gives 

0(2v p ) = 0(v p ) = Pl -p 3 ^O 

while we would expect 0(2v p ) = 20(v p ) if were linear. 

(c) For fixed p = (pi, p 2 > Pa) the mapping is 

(vi, v 2 , v 3 )i ►PsVi + p^, 

which is linear and hence is a 1-form. 

Now 0(U^p)) = 0((1, 0, 0) p ) = p 3 = x 3 (p) and hence 0(U t ) = x 3 ; 

0(U 2 (p)) = 0((O, 1, 0) p ) = pi = Xi(p) and hence 0(U 2 ) = x i; 

0(U 3 (p)) = 0((O, 0, l) p ) = 

and hence = 20(U})dxj = x 3 dxj + X!dx 2 . 



M334 1.5 35 

(d) For fixed p the mapping v p i — ►v p [x 2 + y 2 ] is linear, by Theorem 
1.3.3.(1), and since v p [x 2 + y 2 ] is a real number p is a linear func- 
tional and is a 1-form. Now by the definition of a differential 

v p [x 2 + y 2 ]=d(x 2 + y 2 )(v p ). 

Hence, since 0:v p i — >v p [x 2 + y 2 ] = d(x 2 + y 2 )(v p ), it follows that 
= d(x 2 + y 2 ) = 2x dx+ 2y dy. 

(e) This mapping is linear for each fixed p and so does define a 1-form. 
Obviously it is the zero 1-form, for which again we use the greatly 
overworked symbol 0. 

(f) This fails to five us a 1-form for much the same reason as for part 

<b). 

2. Page 25, Exercise 1(c). 

If \jf = Sfjdxi then ^( Vp ) = 2fi(p) Vi . 
i 

Here i// = (z 2 - l)dx - dy + x 2 dz and hence 

^( v p) = (Pi _1 ) v i " V 2 + P?V3. 
If v p = (1, 2, 3) (0 , _ 2 , l) then tf,(v p ) = -2. 

3. Page 25, Exercise 3. 

0(W) = (x 2 dx - y 2 dz)((xy + yz)\J l - yzU 2 - xyU 3 ) 

= x 2 (xy + yz)dx(Uj) - x 2 yz dx(U 2 ) - x 3 y dx (U 3 ) 

-y 2 (xy + yz)dz(U x ) + y 3 z dz(U 2 ) + xy 3 dz(U 3 ), by linearity, 

= x 2 (xy + yz) + xy 3 = x^ + x 2 yz + xy 3 
= xy(x 2 + xz + y 2 ), 
since dx^) = dz(U 3 ) = 1 and all the other similar terms are zero. 



Page 26, Exercise 5(a). 

By Corollary 1.5.5, df =^£dx + JUdy +_^ldz. 



3x By 9z 



i 



Here f = (x 2 + y 2 + z 2 ) 1 and hence ^1= x(x 2 + y 2 + z 2 j"t 

3x 

Using similar results for and_we obtain 

3y dz 
,i 

df = (x 2 + y 2 + z 2 ) 2 (x dx + y dy + z dz). 



5. Page 26, Exercise 5(b). 

When dealing with the function tan _1 (y/x) we must restrict ourselves to a 
region of E 3 on which the function x is never zero. The largest such domain 
is{pEE 3 : Pl #0}. 



36 M334 1.5 

Now d(tan _1 (y/x)) = (tan" 1 )'(y/x)d(y/x), by Lemma 1.5.7, 

1 



1 + (y/x) 2 
x 2 /l 



•d(y/x), 



* [Idy + yd (1)\ , by Lemma 1.5.6, 

+ v 2 \x x / 



x 2 + y 



_ x 2 /dy_ ydx 



x 2 + y 2 \ x x 2 
x dy - y dx 

x 2 + y 2 



by Lemma 1.5.7, 



6. Page 26, Exercise 6(b). 

By Corollary 1.5.5 

d(xeYZ) = ^fl!> dx + d ±^l dy + S J^1 dz 
3x 3y 9z 

= eV z (dx + xz dy + xy dz). 

Hence df[v p ]= eP*P 3 ( Vl + Pl p 3 v 2 + Pl p 2 v 3 ) and if v p = (1, 2, 3) ( , -2, 1) 
it follows that df [v p ] = e" 2 . 



M334 1.6 37 

1.6 DIFFERENTIAL FORMS 

Introduction 

This section follows on from Section 1.5, where we gave an abstract definition of a 
1-form and then proved that any 1-form is a combination of the differentials of the 
natural coordinate functions. In this section we introduce 0-, 1-, 2- and 3-forms as 
formal expressions in terms of these differentials and define the wedge product and 
exterior derivative of such forms. In Chapter IV and Chapter VI we shall see that 
these higher dimensional forms can be defined in terms of multilinear functionals 
operating on tangent vectors. 

READ: Section 1.6 (pages 26-31). 



Comments 

(i) Page 27: lines 15 to 19 Take these formal expressions as the definitions 

of 0-, 1-, 2-, 3-forms and treat the previous paragraphs as motivation only. 

In certain cases it is possible to simplify the formal expression for a p-form 
without introducing any ambiguity. 

(a) If some, but not all, of the coefficient functions are the zero func- 
tions we can omit the corresponding part of the linear combination. 
For example 

e x dx + xyz dy + Odz = e x dx + xyz dy 

and 

z 2 dx dy + dx dz + dy dz = z 2 dx dy. 

(b) If all the coefficient functions are zero we have the zero p-form and 
when no confusion can arise we just write this as 0. 

That is, 

(i) the constant function with value is the zero 0-form. 

(ii) Odx + Ody + Odz = is the zero 1-form. 

(iii) dx dy + dx dz + dy dz = is the zero 2-form. 

(iv) dx dy dz = is the zero 3-form. 

It is very unlikely that this abuse of the symbol will ever lead to an 
ambiguity. 

(c) If some of the coefficient functions are the constant function with 
value 1 we omit them. For example, 

eY dx + 1 dy + 1 dz = eY dx + dy + dz 

and eY dx dy + dx dy + 1 dy dz = eY dx dy + dy dz. 

(d) If some of the coefficient functions are the constant function with 
value -1 we omit the symbol "1". For example, 

1 dx dy + (- 1) dx dz + Ody dz = dx dy - dx dz 

and (-1) dx dy dz = -dx dy dz. 



38 M334 1.6 

More generally, if some coefficient function is -f, for some more 
commonly used function f, we write ... -f ... instead of 
. . . + (-f) . . . . For example, 

x dx + (-y) dy + dz = x dx - y dy. 

(ii) Page 27: line 23 Using the interpretation of 1 -forms given in the last 

section we can prove that to add 1 -forms or multiply them by some real- 
valued function all we have to do is to add the coefficient functions or 
multiply them by the given function. That is, 

Sfidxj + Sgidxj = 2(fi + gi )dx i5 

f(2fid Xi ) = 2(ffi)dxi. 

Since we have no way of interpreting 2- and 3-forms yet the best we can do 
is to define addition and scalar multiplication in a similar way. That is 

(f ! dx dy + gjdxdz + r^dydz) + (f 2 dxdy + g 2 dxdz + h 2 dydz) 

= (fj + f 2 )dxdy + ( gl + g 2 )dxdz + (hi + h 2 )dydz; 
k(fdxdy + gdxdy + hdydz) = kfdxdy + kgdxdz + khdydz; 
fjdxdydz + f 2 dxdydz = (f t + f 2 )dxdydz; 
g(fdxdydz) = gfdxdydz. 

(iii) Page 27: Example 6.1 This example serves as a definition of the wedge 

product. 

The product is given by the following procedure. 

(a) Expand using the distributive law, bringing the coefficient functions 
to the front, dropping the symbol a but maintaining the order of the 
differentials. 

(b) Drop terms with a repeated differential. 

(c) Reorder the strings of differentials using the alternation rule. 

(iv) Page 28: Definition 6.3 The wedge products dfjAdxj are expanded using 

Corollary 1.5.5. 

That is, 

dfjAdxj = l^h dx! + Oh. dx 2 + Oh. dx 3 ] Adxj 
\dx! 3x 2 dx 3 / 

= i dxjdxj + L dx 2 dxj + L dx 3 dxj. 

bxi 9x 2 9x 3 

The definition of the exterior derivative can be extended to cover 2-forms. If 

17 = f dx dy + g dx dz + h dy dz 
then 

&q = df Adx dy + dgAdx dz + dhAdy dz 
and this expression is again expanded using Corollary 1.5.5. 



M334 1.6 



39 



(v) Page 29: line 13 Though we can multiply 1-forms by any real-valued 

functions, in this formula the coefficients a and b are only real numbers or 
equivalently constant functions. 



(vi) Page 30: lines 11 and 12 Since dx dy dz is obtained from dz dx dy 
by two interchanges the expression is multiplied by (-1) 2 = 1. 



Summary 



Notation 




fdx 


+ gdy + hdz 


Page 27, line 16 


fdx 


dy + gdxdz + hdydz Page 27, line 18 


fdx( 


dy dz 


Page 27, line 19 


0A^ 




Page 27, Example 6.1 


d0 




Page 28, Definition 6.3 


Definitions 




(i) 


Differential form ^ 






0-form 






1-form 
2-form 


Page 27, lines 15-19 




3-form 

) 




(") 


Addition and multip 


lication of forms Page 27, lines 20-23 


(iii) 
(iv) 


Wedge product 0a \jj 
Exterior derivative d< 


Page 27, Example 6.1 

P Page 28, Definition 6.3 and 

Text: page 38 



Results 



(i) The wedge product of 1-forms satisfies the 

alternation rule 0a \]j =^a0 

(ii) The exterior derivative satisfies linear and 
Leibnizian properties 

d(a0 + b\^) = ad0 + bdi// 

d(fg) = dfg + fdg 

d(f0) = dfA0 + fd0 

d(0Al//) = d0Al// — 0Ad^. 



Page 28, Lemma 6.2 



Page 29, line 13 and 
Theorem 6.4. 



40 M334 1.6 

Techniques 

(i) Addition and scalar multiplication of 

differential forms. Page 27, lines 20-23 

(ii) Expansion of the wedge product of 

differential forms. Page 27, Example 6.1 

(iii) Expansion of the exterior derivative. Page 28, Definition 6.3 

(iv) Expansion of the exterior derivatives using 

the linear and Leibnizian properties. Page 29, Theorem 6.4 



Exercises 

Technique (i) 

1. (a) Add the 2-forms 

xdxdy + ydxdz and -xdxdy + e x Ydxdz + 2dydz. 
(b) Multiply the second of the above 2-forms by the function e~ x Y. 

Technique (ii) 

2. Form the wedge product of 

(a) yzdx + dz and dy + zdz, 

(b) yzdx + dz and -xdxdy + e x Ydxdz + 2dydz. 

Technique (iii) 

3. Find the exterior derivative of 

(a) xyz dx + dz, 

(b) xyz dx dy. 

Technique (iv) 

4. Page 31, Exercise 3. 

Theory Exercise 

5. Page 31, Exercise 7. 

Solutions 

1. (a) (xdxdy + ydxdz) + (-xdxdy + e x Ydxdz + 2dydz) 

= (x + (-x))dxdy + (y + e x Y)dxdz + 2dydz 
= dx dy + (y + e x Y)dx dz + 2dy dz 
= (y + e x Y)dx dz + 2dy dz. 



M334 1.6 41 

(b) e" x Y(-xdxdy + e x Ydxdz + 2dydz) 

= -xe- x Ydxdy + e x Ye- x Ydxdz + 2e" x Ydydz 
= -xe'^dx dy + 1 dx dz + 2e~ x Ydy dz 
= -xe- x Ydxdy + dxdz + 2e" x Ydydz. 

2. (a) (yzdx + dz) a (dy + zdz) 

= yz dx dy + yz 2 dx dz + dz dy + z dz dz, by the distributive rule, 

= yz dx dy + yz 2 dx dz + dz dy, since dzdz = 0, 

= yzdxdy + yz 2 dxdz - dydz, by the alternation rule, 

(b) (yzdx + dz) a (-xdxdy + e x Ydxdz + 2dydz) 
= -xyzdxdxdy + yze x Ydxdxdz + 2yzdxdydz 

-x dz dx dy + e x Ydz dx dz f 2dz dy dz, by the distributive law, 

= 2yz dx dy dz + x dz dx dy, since dx dx ay = dx dx dz 

= dz dx dz = dz dy dz = 0, 
= (2yz - x)dx dy dz, since dz dx dy = dx dy dz. 

*• (a) d(xyzdx + dz) = d(xyzdx + ldz) 

= d(xyz) A dx + d(l) a dz 

= d(xyz) A dx, 

since the differential d(l), of the constant func- 
tion with value 1, is the zero 1-form and since 
a dz is the zero 2-form, 

= (yzdx + xzdy + xydz) a dx, 

by Corollary 1.5.5, 
= yzdxdx + xzdydx + xydzdx, 

by the distributive rule, 
= xzdydx + xydzdx, 

since dx dx = 0, 
= -xzdxdy -xydxdz, 

by the alternation rule. 

(b) d(xyzdxdy) = d(xyz) a dxdy 

= (yzdx + xzdy + xydz) a dxdy 

= yz dx dx dy + xz dy dx dy + xy dx dx dy 

= xydz dxdy 

= xydxdydz. 



42 M334 1.6 

4. Page 31, Exercise 3. 

By Corollary 1.5.5, df = 2 JUdxj. 

i 3xj 

By Definition 1.6.3, d(df) = 2 d PI] a dxj. 

i \3xi/ 

Again using Corollary 1.5.5 we have 

d(df) = 2 1 2 32f dx; ) a dxi = 2 2 92f dxjdxj. 
1 \ J a Xi 3xj / J J 3 Xi 3xj 

There are nine terms in this sum, which we must show is zero. We know that 
the terms involving dxidx!, dx 2 dx 2 and dx 3 dx 3 are already zero. The 
remaining six terms occur in pairs such as 

dx 2 dx 1 + dx!dx 2 . 



dxj9x 2 dx 2 3xi 

By the alternation rule this is 

I- + |dxidx 2 . 



\ 9x 2 3x! 3xj3x 2 / 

But we know that 3 2 f/3x 1 dx 2 = 3 2 f/3x 2 3x! for suitably differentiable 
functions, and so the remaining terms in the sum for d(df) cancel in pairs. 

Hence d(df) = 0. 

By Theorem 1.6.4, 

d(fdg) = df a dg + fd(dg), 

= df a dg, by the above result. 

Page 31, Exercise 7. 

Any 1-form is of the form 

= 2 fjdxj, 
i 

where f j and xj are differentiable functions. 
By Definition 1.6.3, 

d0 = 2 df} a dx} 
i 

and by Theorem 1.6.4 (3), 

d(d0) = 2 (d(dfi) a dxj - df} a d(dxi)). 
i 

By Exercise 3 we know that both d(dfj) and d(dxj) are zero and so 
d(d0) = 0. 



M334 1 1.7 43 

1.7 MAPPINGS 

Introduction 

This section follows on from Sections 1.3 and 1.4. 

In Section 1.3 we introduced directional derivatives in order to differentiate func- 
tions from E 3 to R, and in Section 1.4 we introduced the componentwise differen- 
tiation of functions from R to E 3 . In this section we combine these ideas in order to 
differentiate functions from E n to E m , for any positive integers n and m. The 
derivative mappings of these functions turn out to be linear transformations between 
corresponding tangent spaces and we shall describe the matrices representing them. 

READ: Section 1.7 (pages 32-39). 

Comments 

(l) Section 1.7 In this section there are several straightforward general- 

izations of definitions that have previously been applied to E 3 only. Those 
that are not specifically mentioned are: 

(a) Natural coordinate functions x 1} . . . , x n , which generalize those of 
Definition 1.1.2. If p is a point in E n then xj(p) = pj (i = 1, 2, . . . , n). 

(b) Tangent vectors v p in E n , which generalize those of Definition 1.2.1. 
The tangent vector v p represents the "arrow" from p to p + v in E n . 

(c) Curves a(t) = (a^t), . . . , a n (t)) in E*\ which generalize those of 
Definition 1.4.1. 

(d) Straight lines 1 1 — ► p + tv in E n . 

(e) The velocity vector a'(t) = ((d ai /dt)(t), . . . , (da n /dt)(t)) a(t) for a 
curve a in E n , which generalizes Definition 1.4.3. 

(f) The directional derivative v p [f] of a function f on E n with respect to 
a tangent vector v p , which generalizes that of Definition 1.3.1. That 
is>v p [f] =(d/dt)(f(p + tv))| t = . 

(ii) Page 36: line 3 

F*(v) = F( p + tv) r i(0) 

= ((Pi + tv 1 ) 2 -(p 2 +tv 2 ) 2 ,2(p 1 +t Vl )(p 2 +tv 2 ))'(0) 

= (2( Pl + tw l )v 1 -2( Pl +tv 2 )v 2 ,2( Pl +tv,)v 2 +2 Vl (p 2 +tv 2 ))(0) 

= 2(p!V! - p 2 v 2 , pjv 2 + p 2Vl ). 

(iii) Page 37: following Corollary 7. 7 To obtain the j-th column of a matrix 

representing a linear transformation with respect to given bases, we find the 
coordinates for the image of the j-th basis vector in the domain, in terms of 
the basis for the codomain. 



44 M334 1.7 

In Example 7.3(2) we take as basis vectors for the domain U^p) and U 2 (p) 
and for the codomain Uj(F(p)) and U 2 (F(p)). Hence, by Corollary 7.7, the 
linear transformation is represented by 



lili(p) 2!i(p)\ 

9u 9v 



t 3u dv / 



where f j = u 2 - v 2 and f 2 = 2uv. 

All such matrices are called Jacobian matrices. 

Hence at p = (pi ,p 2 ) the Jacobian matrix is 

/2u(p) -2v(p)\ /2pi -2p 2 ^ 

\2v(p) 2u(p)/ \2 P2 2p u 

The image of a typical vector v = (v x ,v 2 ) under this transformation is 




which is the answer obtained on page 36, line 3. Note that while dealing with 
matrices we represent tangent vectors by column vectors. 



Additional Text 

(a) Let f be a mapping from R to R. Then the Jacobian matrix is the 

1 X 1 matrix (df/dt), where here we do not need partial derivatives since f is 
a function of t only. 

Now a tangent vector in R, v p is an arrow from p to p + v, 



v 



p + V 



and this is mapped by f^p to the tangent vector v(df/dt)(p) at f(p). That is it 
is multiplied in length by (df/dt) (p). 

df, , 
v (p) 

dt 

» ► •— 



f( P ) f(p)+vii( P ) 

dt 



M334 1.7 



45 



(b) Suppose we have a curve a : R ► R 3 , where a has coordinate functions 

(«!, a 2 , a 3 ), i.e. 

a : t .— > a(t) = (^(t), a 2 (t), a 3 (t)). 
Then we obtain the 3X1 Jacobian matrix 

dt 
da, 



dt 




A tangent vector v, at p, is mapped to the tangent vector a* p (v), at a(p). 
Reverting to row vectors, this is the tangent vector 



vi^( P ), vi^( P ), v^( P) ) 

dt dt dt / a(p) 




(c) Suppose we have a real-valued function F : E 3 >R, then the Jacobian 

matrix is the 1 X 3 matrix 

/3F dF 3F\ 
\3x, dxj 3x 3 i 



If v p - v 2 is a typical tangent vector, it is mapped by F^ 



to the tangent vector 



/ 3F , . 3F , . 3F , A 

(P)» (p)» (p) /vA at F(p) in R. 

\ 3x t 3x 2 3x 3 / / \ 



v 2 

, V 3 



46 M334 1.7 

This is V! (3F/3x 1 )(p) + v 2 (3F/8x 2 )(p) + v 3 (3F/3x 3 )(p), which we have 
already met as the directional derivative v p [F] . 



p + v 



| » 3F . ., 3F , x , 3F , , 

1 V 1T" (P) +v 2— (P) +V 3T- (P) 

OXj ox 2 ox 3 

F(p) F(p)+v.VF(p) 

(d) The Composite Rule Suppose we have two mappings F : E n ► E m and 

G : E m ► E r , then we can form the composite mapping G ° F : E n ► E r . 

The obvious question is whether we can calculate the derivative map 
(G o F)* in terms of the derivatives G* and F*. The answer is yes. We state 
the result, which should be intuitively obvious, and investigate some of its 
consequences. 

Theorem 

For mappings F : E n ► E m and G : E m ► E r 

(G oF), = G* o F*. 

To be more precise we need to take into consideration the point at which each of 
the derivatives is evaluated. 

Now 

F* p :T p (En) — T F(p) (Em) 

G*F(p) : T F (p)(E m )— ^T G oF(p)(E r ) 
(G o F)^ : T p (En) — . T G F(p) (E') 

and hence (Go F) 5|ep = G^p) o F^ p . (This may remind you of the composite rule 
for ordinary differentiation — D(g p f) = (Dg ° f).Df .) 

Considering all points p of the domain at the same time we can write the composite 
rule as 

(GoF) Jle = G* F oF*. 

This form of the composite rule accounts for all of the composite rules that we have 
encountered. When we expand it in terms of Jacobian matrices the composition on 
the right hand side then stands for the usual matrix multiplication. 



M334I.7 



47 



(i) Given maps f : R ► R and g : R ► R, the cgmposite rule gives the form 

of the chain rule encountered in Ml 00. 

(ii) Given a map h : R ► R followed by a : R ► E 3 , as in Lemma 1.4.5, the 

composite rule gives 



(iii) 



(a°h)*s = a*h(s) oh * s - 

Now h, = and a, is represented by 

ds 



dt 



AOLr 



dt 
doc 3 



\dt 
Hence if we write ft = a o h the composite rule gives 

/*(.)\ . /^(h(s))\ .dh (s) 
/ dt /dt I ds 



d£ 
dt 



Ms) 



(Ms)) 



dt 



dt 



</ I 



*(.)/ \*S(h(.))| 



dt 



dh 



i.e.0'(s)=.^i(s)a'(s). 
ds 

Given a map f : E 3 ► R followed by h : R 

write the composite rule as 

(h°f)* = h* f of,. 



•+R, as in Lemma 1.5.7, we can 



Now h* is just the derivative which can be written h', and so h* f = h'(f). 
Hence in terms of Jacobian matrices 

[ 9 (Mf)) j 3(h(f)) ] 3(h(f)) \ = h , (f) / 3f^ £_ af\ 

» 3x i 9x 2 ^x 3 / ^a Xl a X2 ' a x l' 

i.e. WL) = h'(f)lL 0=1,2,3). 
3xj dxj 



(iv) Given a map a : R ► E 3 followed by a map f : E 3 

then the composite rule states 



■* R, as in Lemma 1.4.6, 



( foO 0*t = f *a(t) oOf *f 



48 



M334 1.7 



In terms of Jacobianjnatrices, this becomes 

?m ( t) =(!L ( «(t)), 5. («(,)), ^_ Wt ))) /i«L(t)\ 

dt iBx! 8x 2 3x 3 // dt 



so 




dt i=l 3xj dt 

Now read the proof of Lemma 1.4.6 and Lemma 1.3.2, where 
a : 1 1 — ► p + tv, so that (dcq/dt)(0) =(d/dt)(pj + tv;) | t = q vj. 



Summary 




Notation 


F = (fi,...,f m ) 


F* 


Ibi df\ 
8u 8v 




i 9 g ^g 

\3u a7/ 




Definitions 


(i) Euclidean coordinate functions (f l5 . 


(ii) Image of curve under a mapping F(a) 


(iii) The derivative map F* 


(iv) Jacobian matrix 


Results 


(i) The 


t derivative map can be describe 



.fm) 



the coordinate functions and directional derivatives 

F*(v) = (v[f,],...,v[f m ]). 

(ii) The derivative map F^p : T p (E n ) >T F (p) (E m ) 

is a linear transformation. 

(iii) The effect of F* on the standard basis is 

F*(Uj) = 2^L tV 
i 9x; 



Page 33, Definition 7.1 
Page 35, Definition 7.4 



Page 37 and Text, page 44 



Page 33, Definition 7.1 
Page 33, Definition 7.2 
Page 35, Definition 7.4 
Page 37 and Text, page 44 



Page 36, Theorem 7.5 



Page 36, Corollary 7.6 



Page 37, Corollary 7.7 



M334 1.7 49 

(iv) The derivative map preserves velocities. 

If = F(a) then 0' = F # (a'). Page 38, Theorem 7.8 

(v) The composite rule: 

If F : E n >E m and G : E m >E r then 

(G° F )*p = G*F(p)°V 

Techniques 

(i) Description of mappings in terms of Euclidean co- 

ordinate functions. Page 33, Definition 7.1 

(ii) Pictorial description of simple mappings. 

(iii) Determination of derivative maps, 

(a) from first principles Page 35, Definition 7.4 

(b) from directional derivatives of coordinate 

functions Page 36, Theorem 7.5 

(c) from Jacobian matrix. Text, page 44 

(iv) Calculation of the derivative mapping of a 
composite using the composite rule: 

(G°F)* = G* F °F* 

where if the derivatives are represented by Jacobian 
matrices the composition on the right is matrix 
multiplication. 



Exercises 

Technique (i) 

1. Express the following mappings from E 3 to E 3 in terms of Euclidean 
coordinate functions 

(a) F:p.— ►-3p, 

(b) F: P h— ► (ePiP2,p3+2 Pl ,pJ). 

Technique (ii) 

2. Page 39, Exercise 2. 

Technique (iii) 

3. (a) Page 39, Exercise 4. 

(b) Find F*(v p ), using directional derivatives, for F = (x cosy, x siny, z), 
v = (2,-l, 3)andp = (0, 0, 0). 

(c) Describe the derivative mappings for F given in Example (1), page 34. 



50 
Technique (iv) 



M334 1.7 



4. Use the composite rule for Jacobian matrices to find (9f/9x, 9f/9y, 9f/9z) 

for the composite mapping in Exercise 4(a), page 6. 



Solutions 



(a) Since p • — ► - 3p, 

F(p) = (-3pi, -3p 2 , -3p 3 ) 

= (-3x(p), -3y(p), -3z(p)) 
and so, in terms of Euclidean coordinate functions, 

F = (-3x, -3y, -3z). 

(b) F(p) = (ePiP2, p 3 +2p lf p\) 

= ( e x(p)y(p), z(p) + 2x(p), (x(p)) 2 ) 
= (exy( p ),(z + 2x)(p),x 2 (p)) 

and so F = (e x Y, z + 2x, x 2 ). 



Page 39, Exercise 2. 

The lines u = 1, and v = 1 are the sets of points p such that u(p) = Pi = 1 and 

v(p) = p 2 =1. These are shown on the following diagram. 





V 


u= 1 
















(1.1) 


v= 1 


(0,0) 






► 



A point on the horizontal line is of the form (t, 1) and is mapped by F to 
(t 2 - 1, 2t). So, for instance, (0, 1) i — >(-l, 0), (1, 1) " — >(0, 2) and 
(-1, 1)h->(0,-2). 



M334 1.7 



51 



(-1,0) 




► u 



How do we describe such a curve? A point p= (pi,p2) is on the curve if 
p t = t 2 - 1, p 2 = 2t for some t. From these identities we deduce that 
4(pj + 1) = p 2 . Using the notation of Section 1.4 we write 

pGM: 4(u+lj- v 2 =0. 

The line u = 1 consists of points of the form (1, t). These are mapped by 
F to (1 - t 2 , 2t). This is a typical point on the parabola v 2 = -4(u - 1), which 
is shown on the following diagram. 




► u 



3. (a) Page 39, Exercise 4(a). 



F*(v p )=_(F( P + tv))| t = atF( P ) 
dt 



= ( F (Pl + tV!, p 2 +tV 2 , p 3 +tV 3 ))| t = 

dt ' 

= — (Pi +tv i "P2 -tv 2 ,p! +tvj +p 2 +tv 2 ,2p 3 +2tv 3 )l t = o 

dt ' 



= (vi -v 2 , Vl +v 2 , 2v 3 ) at F(p). 



52 



M334 1.7 



Page 39, Exercise 4(b). 

If F is linear then F(p + tv) = F(p) + tF(v). 

Hen <* F*( v ) = l(F(p + tv)) | t =0> ^ F(p), 
dt 

=l(F(p) + tF(v))| t=0 ,atF(p), 
dt 

= F(v) at F(p), 

so F^Cvp) = F(v) F ( p ). 

(b) If F = (x cos y, x sin y, z ), then by Theorem 1.7.5 
F*(v p ) = (v p [xcosy], v p [xsiny], v p [z]) F ( p ). 
Now if v = (2,-1,3) and p = (0,0,0), then by Lemma 1.3.2 

/\ rl B 

v p [xcosy] = 2 (xcosy)(p) - — (xcosy)(p)+ 3 — (xcosy)(p) 

dx 9y 3z 

= 2 cosy (0,0,0) + x sin y (0,0,0) + 3.0(0,0,0) 
= 2cos0 + OsinO + 3.0 
= 2. 

Similarly, 

v p [xsiny] = 2 sin y(0,0,0) - xcosy(0,0,0) + 

= 2sin0 - OcosO = 0. 
v p [z] = 2,0 - 1.0 + 3.1 
* 3. 
So 

F*(v p ) = (2,0,0)( 0cos0>0 sin0, 0) = (2,0,3) (0,0,0)- 



(c) If F = (x - y, x + y, z) the Jacobian matrix for F^ is 



-1 



Hi 


dfi 


Mi 


3x 


a y 


dz 


3f 2 


3f 2 


8f 2 


3x 


3y" 


3z 


df 3 


3f 3 


M3 


8x 


8y 


3z 



This maps the vector v to 
/ 1 -1 



1 

\ 



' V1 \ 


vi -v 2 \ 


v 2 = 


Vl +v 2 


\ v 3 / 


\ v 3 



andsoF*(v p ) = (v t -v 2 ,V! + v 2 , v 3 ) F ( p ), as expected. 



M334 1.7 



53 



4. We have f described as a composite of functions, f = h o g, where 

g : p i ► (gi(p), g 2 (p), g3(p))- That is, g u g 2 , g 3 are the coordinate functions 

for g and so in this example 

g = (x + y, y 2 , x + z). 



Since f : E 3 ► R, g : E 3 ► E 3 and h : E 3 ► R, the corresponding 



I : iL" ► K, 


g:IV 


► 


m matnces are 




(3f 9f 


9f\ 




r 


Ox 9y 


9z/ 


dgi 9gi 


3gi\ 


dx 9y 


9z 




a g2 3g 2 


9g2 




9x 9y 


9z 





2y 



a g3 dga 9g 3 
9x 9y 9z 



, /9h 9h 9h\ /0 
and [_, — , _q = (2x, -z, -y). 



; 9x 9y 9zJ 
Now by the composite rule f* = h* / \ © g^ 



so 



i£(p), i£(p),H<p)U( 2x (g(p)), -z(g( P )) ( -y(g( P ))) 

9x 9v 9z / » / 



1 



2y(p) 



1 



(2gi(p),-g3(p),-g 2 (p)) l l 1 ° 

2p 2 

1 1 

(2pi + 2p 2 ,- Pl -p3,-p 2 2 ) /l 1 

2p 2 

1 1 

= (2 Pl + 2p 2 -pJ,-2 Pl + 2p 2 -2 Pl p 2 -2p 2 p 3 , -pj), 
and hence 

/9f 9f 9f\ /0 

— > , — = (2x + 2y - y 2 , 2x + 2y - 2xy - 2yz, -y 2 ). 

\3x 9y 9z/ 



54 M334 1 

FURTHER EXERCISES AND SOLUTIONS 

Section 1. 1 

Technique Exercises 

Pages 5-6, Exercises 1(b), 2(a), 2(d), 4(b), 4(c). 

Solutions 

1(b). xy(2y + x)sin z 

2(a). 



2(d). t 4 (l-t 3 ) 




4(b). — = 2e2x(i. 
3x 


-eV) 


4(c). ii = 4x. 
3x 




Section 1.2 





Technique Exercise 
Page 11, Exercise 4. 

Other Recommended Exercise 

Page 11, Exercise 5. This extends the concepts of linear independence and linear 
combinations to cover vector fields. 

Solutions 

4. f = kx 2 , g = -ky 2 for any real function k on R 3 , 

e.g. f = x 2 , g = -y 2 . 
5(a). V 1 (p) = (l,0,-p 1 ) p 

V 2 (p) = (0,l,0) p 

V 3 (p) = (Pi,0,l) p 



and since 



1 





Pi 





1 





Pi 





1 



1 + p? =£ these vector fields are linearly independent. 



5(b). (xU! + yU 2 + zU 3 )(p) = (pi,p 2 ,p 3 )p 

. PiUzPs) Vl (P) + P2 v 2 ( P ) + (pLLElV, (p) 

(1+Pl) (1 + Pl) 



and hence 



xu t + y u 2 +zu 3 = x(1 ~ z) Vi +yv 2 + (x +z) v 3 . 

(1+x 2 ) (1+x 2 ) 



M334 1 55 

Section 1.3 

Technique Exercises 

Pages 14-15, Exercises 1(a), (b), 2(a), (b), 3(a)-(d), (f). 

Solutions 

1(a) and 2(a). 

1(b) and 2(b). 896 

3(a). y 3 

3(b). -3xz 2 

3(c). yz 2 (y 2 z-3x 2 ) 

3(d). -yz 2 (y 2 z-3x 2 ) 

3(f). 

Section 1.4 

Technique Exercise 
Page 21, Exercise 8. 

Other Recommended Exercises 

Page 21, Exercise 4. This exercise shows that a curve is determined uniquely by its 
initial point and velocity vectors. 

Page 21, Exercise 6. This theory exercise deals with the definition of the directional 
derivative. 

Page 21, Exercise 9. This deals with the geometrical interpretation of tangent lines. 
Solutions 

8. /3(s) = |s,2,V21ogs), 

0'(s) = (l,-±,V£\ 

a'(h(s)) = /s,-i,V2J. 

dh, x 1 au 

— (s) =_and hence j3'(s) = ™( s ).a'(h(s)). 

ds s ds 

4. a(t) = (l + l\il,et-6). 

3 2 



56 M334 1 

6. Let a be any curve with ol (0) = v p . 

Thenfi!^l(0) = a(0)[f] = v p [f]. 
dt 

9. (a) u i — ► (2, 2u, u). 

(b) u h-» (V2(l - u), V2(l + u),JL+ u). 



Section 1.5 

Technique Exercises 

Pages 25-26, Exercises 1(a), (b), 3(a), (c), 4, 6(a), (c). 

Other Recommended Exercises 

Page 25, Exercise 2. This expresses the operation of a 1-form on a vector field in 
terms of Euclidean coordinate functions. 

Page 26, Exercise 8. This deals with the Leibnizian property of d. 

Page 26, Exercises 9, 10. These explore the relationship between the differential of a 
function and its maxima and minima. 

Page 26, Exercise 11. This deals with the relationship between the differential and 
the difference operator. 

Solutions 

1(a). 4 
1(b). -4 

3(a). x 3 +x 2 (l +z)-xy 2 -LI 



x 



4(a). 5f 4 df 
df 



4(b). 



4(c). 



2>/f 
2fdf 



1+f 2 

6(a). df = y 2 dx + (2xy - z 2 )dy -2yz dz, -10. 

6(c). df = cos(xy)cos(xz)(y dx + x dy) - sin(xy)sin(xz)(z dx + x dz), -2. 

0(V) = 22 f iV ; dxi(U;) = 22 fjv^i; = 2 f iVi . 
i j J J i j J J i 

8- d(fg).(v p ) = v p [fg] = v p [f] g + fv p [g] 

=df(v p )g + fdg(v p ) 

= (gdf + fdg)( Vp ) 



2 



M334 1 57 

and hence d(fg) = g df + f dg. 

9. df(v p ) =-* (pK +.21 (p)v 2 +iL(p)v 3 

3x 3y 3z 

and hence df(v p ) = for all v if and only if 

i£(p)-i!(p)>i£(p)-o. 

3x 3y 3z 

Here Jl = -2xy.il = 1 - x 2 - 2yz,J! = 1 - y 2 
3x 3y 3 Z 

and the critical points are ± (0, 1, 5). 

10. If p is a local maximum or minimum then t = is a local maximum or 
minimum of the function 1 1 — ► f(p + tv). 

Hence df [v p ] = v p [f] =_1 (f(p + tv)) | t=Q = . 

dt 

11(a). By Taylor's Theorem 

f(p + tv) = f(p) + t£(f(p + tv)) | t=0 + Remainder 
dt 

= f(p)+tdf(v p )+R. 

Hence f(p + v) - f(p) ~ df(v p ). 



11(b). Exact: -0-420 



Approximate: df(v p ) = 1*?* dx +JL dy - £l dz) (-0-1, 0-1, 0.2) (1> 1 . 5> x) 



2 



Section 1.6 

Technique Exercises 

Page 31, Exercises 1, 2, 4(a), (b), (d), 6. 

Other Recommended Exercises 

fhTd.?' ExerC r/; ^ "P*"" th e wedge product of three 1-forms in terms of 
the determinant of their Euclidean coordinates. 

Page 31, Exercise 8. This exercise deals with the relationship between the exterior 
derivative and the vector operations div, grad and curl. 

Solutions 

1(a). yz cosz dx dy - sinz dx dz - cosz dy dz, 
sinz dx dy + z sinz dx dz + z cosz dy dz, 
-yz dx dy - yz 2 dx dz + dy dz. 



58 M334 1 

1(b). -z dx dy - y dx dz, 

-cos z dx dz + sinz dy dz, 

0. 
2. d0 = dx dy, d\}/ = - dy dz, 

y 2 

0a \p = dx dy and 

y 

d(0Ai//) = d0A^ - 0Adi|/ = dx dy dz. 

y 

4(a). d(fdg + gdf) = d(d(fg)) = 0. 

4(b). (df - dg)A(df + dg) = 2 dfAdg. 

4(d). (l-f)dfAdg 

6. dx dy dz = r dr d& dz 

5. Expand both the determinant and the triple wedge product in terms of the 

Euclidean coordinate functions. Both sides give 

(f 11^33 ~ fllf 2 3f32 + ^23131 " ^2^33 + f 1^2^32 " * 13^22^1 )dx dy dz. 

8(a). df = zfl dxi t W, 2^1 Ui = grad f. 
9x • 3xj 

8(b). If V = Sfi Ui then = Sfj dxj, 

d =(^ -!M d XldX2 + (2f! -!M d Xl dx 3 + [Hl -^)dx 2 dx 3 

and d0 < ( 2 ) , curl V. 
8(c). If V = 2fiUjthenT7 = f 3 dx 1 dx 2 - f 2 dxjdx 3 + f!dx 2 dx 3 . 

dr?=Fi +^!i +^1\ dx, dx 2 dx 3 and hence dr? ( ( 2 ) , (div V)dx dy dz. 
\dxi 9x 2 3x 3 ; 



Section 1.7 

Technique Exercise 
Page 39, Exercise 1. 

Other Recommended Exercises 

Page 40, Exercises 5 and 6. These describe the derivative of a mapping in terms of 
the differentials of its Euclidean coordinate functions. 

Page 40, Exercise 8. This deals with the definition of the derivative. 

Page 40, Exercise 9. The derivative mapping preserves directional derivatives. 



M334 1 59 

Page 40, Exercise 11. This deals with inverse mappings, which you can read about in 
Exercise 10. 

Page 40, Exercises 7 and 12. These deal with the derivative of a composite mapping. 



Solutions 

1(a). (0, 0). 

1(b). (-3,1), (3,-1). 

1(c). (0,0), (1,0). 

5. F^ = (cos y dx - x sin y dy, sin y dx + x cos y dy, dz) 

(a) (2,0,3) (0) o,0) 

(b) (2,2,3) (0 ,2,7T) 

6. The Jacobian matrix at p is 
/ cos p 2 sin p 2 

-p 1 siwp 2 piCosp 2 

,0 1 

This has determinant pj and hence F is not regular at points for which 
Pi=0. 

Let a be any curve such that a (0) = v p , then if )3 = F(a) it follows that 
j8 , (0) = F*(a , (0)) = F«(v p ). 

Let a : 1 1 ► p + tq. 

R-^E n iE m _l,R 

d 



v p [g(F)] = a'(0)[g(F)] = -^(g(F( a )))| t=0 . 

dt 
F*(v p ) = (F(a))'(0). 

F*(v p )[g] =(F(*))'(0)[g] =l(g(F(<*)))| t=0 . 

dt 

Hence v p [g(F)] =F*(v p )[g]. 

11(a). F- 1 = (v,ue" v ). This is a diffeomorphism. 

-L JL 

11(b). F" 1 = (u , v + u ). This is not a diffeomorphism since it is not differentiable 
when u = 0. 

11(c). F" 1 = ( ( " u ~ 2v ) , 5 - u - v|. This is a diffeomorphism. 



7- GF = (g 1 (f 1 ,f 2 ), g2 (f 1 ,f 2 )). 

12(a). GF = (g 1 (f 1 ,...,f m ),...,g p (f 1> ..., fm)) 



60 M334 1 

12(b). (GF)*(cO =? ((GF)(a))\ by Theorem 1.7.8, 

=(G(F(*)))' 

=G*((F(c*))') 

=G*F*(c*'). 

12(c). Since any v p = a'(0) for some a it follows that (GF)* = G*F*. 



M334 I 61 

DIFFERENTIAL GEOMETRY 

I Calculus on Euclidean Space 

II Frame Fields 

III Euclidean Geometry 

IV Calculus on a Surface 

V Shape Operators 

VI Geometry of Surfaces in E 3 



1§< PA^TII THE OPEN UNIVERSITY 

! 

I Mathematics: A Third Level Course 



9 



DIP 

PART II FRAME FIELDS 



DIFFERENTIAL GEOMETRY 



9 



THE OPEN UNIVERSITY 

Mathematics: A Third Level Course 

DIFFERENTIAL GEOMETRY PART II 



FRAME FIELDS 



O.U. COURSE U NITS COLLECTION 
FOR REFERE^'cToNLY 



A commentary on Chapter II of O'Neill's 

Elementary Differential Geometry 
Prepared by the Course Team 



THE OPEN UNIVERSITY PRESS 



Course Team 



Chairman: 



Mr. P.E.D. Strain 
Dr. R.A. Bailey 



Lecturer in Mathematics 
Course Assistant in Mathematics 



With assistance from: 



Dr. J.M. 
Mr. G.J. 
Dr. P.M. 
Mr. P.B. 
Dr. F.C. 
Mr. T.C. 
Mr. R.J. 
Dr. C.A. 
Mr. M.G 



Aldous 
Burt 
Clark 
Cox 

Holroyd 
Lister 
Margolis 
Rowley 
. Simpson 



Senior Lecturer in Mathematics 
Lecturer in Educational Technology 
Lecturer in Physics 
Student Computing Service 
Lecturer in Mathematics 
Staff Tutor in Mathematics 
Staff Tutor in Mathematics 
Course Assistant in Mathematics 
Course Assistant in Mathematics 



Consultants: 



Prof. S. Robertson 
Prof. T. Willmore 



Professor of Pure Mathematics, 
University of Southampton 

Professor of Pure Mathematics, 
University of Durham 



4AH CXMM. ^o, 

4- 




The Open University Press, Walton Hall, Milton Keynes 

First published 1975 

Copyright © 1975 The Open University 

All rights reserved. No part of this work may be reproduced in any form, by mimeograph or any other means 
without permission in writing from the publishers. 

Produced in Great Britain by 
The Open University Press 

ISBN 335 05700 4 

This text forms part of the correspondence element of an Open University Third Level Course. The complete 

list of parts in the course is given at the end of this text. 

For general availability of supporting material referred to in this text, please write to the Director of 

Marketing, The Open University, P.O. Box 81, Milton Keynes, MK7 6AT. 

Further information on Open University courses may be obtained from The Admissions Office, The Open 
University, P.O. Box 48, Milton Keynes, MK7 6AB. 



1.1 



M334TI 



CONTENTS Page 

Set Book 4 

Bibliography 4 

Conventions 4 

II. 1 Introduction and Dot Product 5 

II.2 Curves 16 

II. 3 The Frenet Formulas 28 

II.4 Arbitrary-Speed Curves 38 

II. 5 Covariant Derivatives 49 

II. 6 Frame Fields 55 

II. 7 Connection Forms 59 

II. 8 The Structural Equations 61 

Partial Differentiation 66 

FurtherXxercises and Solutions 72 



4 M334 II 

Set Book 

Barrett O'Neill: Elementary Differential Geometry (Academic Press, 1966). It is 
essential to have this book: the course is based on it and will not make sense without 
it. 



Bibliography 

The set books for M201, M231 and MST 282 are referred to occasionally; they are 
useful but not essential. They are: 

t).L. Kreider, R.G. Kuller, D.R. Ostberg and F.W. Perkins: An Introduction to 
Linear Analysis (Addison-Wesley, 1966). 

E.D. Nering: Linear Algebra and Matrix Theory (John Wiley, 1970). 

M. Spivak: Calculus, paperback edition, (W.A. Benjamin/Addison-Wesley 1973). 

R.C. Smith and f. Smith: Mechanics, SI edition (John Wiley, 1972). 

Also mentioned in this Part is: 

T. Wilmore: An Introduction To Differential Geometry (O.U.P., 1964). 



Conventions 

Before starting work on this text, please read M334 Part Zero. Consult the Errata 
List and the Stop Press and make any necessary alterations for this chapter in the set 
book. 

Unreferenced pages and sections denote the set book. Otherwise 

O'Neill denotes the set book; 

Text denotes the correspondence text; 

KKOP denotes An Introduction to Linear Analysis by D.L. Kreider, R.G. 
Kuller, D.R. Ostberg and F.W. Perkins; 

Nering denotes Linear Algebra and Matrix Theory by E.D. Nering; 

Spivak denotes Calculus by M. Spivak; 

Smith denotes Mechanics by R.C. Smith and P. Smith. 

References to Open University Courses in Mathematics take the form: 
Unit Ml 00 22, Linear Algebra I 
Unit MST 281 10, Taylor Approximation 
Unit M201 16, Euclidean Spaces I: Inner Products 
Unit M231 2, Functions and Graphs 
Unit MST 282 1, Some Basic Tools. 



M334 II. 1 



II. 1 INTRODUCTION AND DOT PRODUCT 



Introduction 

This whole section depends only on Sections 1.1 and 1.2, and may be read easily 
even if you are not entirely confident about the later sections of Chapter I. 

It provides a revision of the basic vector concepts that you have met in MST 282 and 
M201. You should make sure that you are familiar with the following ideas: the dot 
product, cross product, triple scalar product and norm for E 3 (Unit MST 282 1, 
Some Basic Tools, Section 2); the dot product, norm, Schwarz inequality, angle, 
orthogonality, orthonormal bases and orthonormal expansions for general Euclidean 
vector spaces (Unit M201 16, Euclidean Spaces I: Inner Products); orthogonal 
matrices (Unit M201 24, Orthogonal and Symmetric Transformations, Section 1.2). 
You will also find it useful to revise the technique of evaluation of a 3 X 3 deter- 
minant (KKOP, page 684). 

These vector ideas are revised and then applied to the tangent vector spaces Tp(E 3 ). 
For example, the dot product of two tangent vectors with the same point of appli- 
cation is defined by 

v -w = v*w. 
v p w p 

We find that the usual vector space results carry over to each of the tangent spaces. 
Later on we shall extend these definitions to vector fields by the pointwise 
principle: for example, if V and W are vector fields the dot product of V and W, 
V'W, will be the function from E 3 to R whose value at p is V(p)'W(p). 



READ: Introduction to Chapter II and Section II. 1 (pages 42-48). 



Comment 

(i) Page 47: the determinant and cross product The determinant is a 

function from the set of all square real matrices into the real numbers. It is 
written by replacing the brackets of the matrix by straight lines. The rules 
for evaluating it on 2 X 2 and 3X3 matrices are as follows: 



a ll a 12 

a 21 a 22 



a ll a 22 - a 21 a 12 



a ll 



a 21 



■■12 



L 13 



a 22 a 23 



a 31 a 32 a 33 



= a ll a 22 a 33 +a 21 a 32 a 13 +a 31 a 12 a 23 

- a 31 a 2 2a 13 - a2ia 12 a 3 3 - a n a 32 a 23 
(see KKOP, page 684) 



which may be written as 



a n a i2 



L 13 



a 21 a 22 a 23 

a 31 a 32 a 33 



= a 



11 



a 22 a 23 

a 32 a 33 



+ a 



12 



+ a 



13 



M334 II. 1 



a 23 a 21 

a 33 a 31 

a 21 a 22 

a 31 a 32 



v 2 


v 3 


+ U 2 (p) 


v 3 


Vl 


+ u 3 ( P ) 


Vl 


V 2 


w 2 


w 3 




w 3 


w 2 




Wj 


w 2 



This last expression gives the expansion of the formal determinant for v X w 
as 



U!( P ) 



= (v 2 w 3 -v 3 w 2 ) U!(p) + (v 3 wx - v 2 w 3 ) U 2 (p) + (v,w 2 -V 2 W!) U 3 (p) 

= (v 2 w 3 - v 3 w 2 ,v 3 w 1 -v 1 w 3 , Vl w 2 - V 2 W!) p . 

O'Neill assumes that you are familiar with the following properties of deter- 
minants: 

(a) a determinant* is linear as a function of each of its rows; 

(b) interchanging two rows of a determinant multiplies its values by - 1; 

(c) a determinant has value zero if and only if its rows are linearly 
dependent vectors. 

In Unit M 201 5, Determinants and Eigenvalues, you learnt these properties 
for the columns of a determinant. However, in Section 1.5 of that unit you 
saw that the determinant of the transpose of a matrix is the same as the 
determinant of the original matrix, and thus all these results are true for 
rows as well. 

By Lemma II. 1.8, v X w is orthogonal to both v and w, and 

H-v X w || 2 = (vv) (ww) - (vw) 2 
= Hv|| 2 ||w|| 2 -(vw) 2 . 
The angle # between v and w is defined by 

v*w= ||v||||w|| COS # 
(see page 44), and so 

llv X w|| 2 = ||v|| 2 ||w|| 2 - (||v|| Hwll cos I?) 2 
= llv|| 2 ||w|| 2 (l-cos 2 #) 
= llv|| 2 ||w|| 2 sin 2 #. 



M334 III 

Taking square roots, 

llvXwII =||v||||w||(sin 2 0) 2 . 



1 

.2 A x2 



For any given value of cos # (in [- 1,1] ) we can choose # to lie in the range 
< # < 7T, that is # is the smaller angle between v and w; for # in this range 
sin # is non-negative and so (sin 2 dp = sin &. Thus 



Iv X wll = llvllllwll sin # 



and so 



v X w = ||v||||w|| sin # n 
where n is a unit vector orthogonal to both v and w. 



v X w 




► w 




In fact n is obtained from v and w by the right-hand rule (page 48, penulti- 
mate paragraph), and so O'Neill's definition of cross-product is equivalent to 
that given in Unit MST 282 1, Some Basic Tools. 

One consequence of the definition of cross-product is the following. 

Lemma II. 1. A If c 1 and e 2 are orthogonal unit tangent vectors and 

e 3 = e x X e 2 then e 3 is a unit tangent vector orthogonal to both e! and e 2 
and 



e i X e 2 = e 3 ; 
e 2 X e 3 = ej ; 
e 3 X ei = e 2 ; 



e 2 X e l = -e 3 ; 
e 3 X e 2 = -e x ; 
e i X e 3 = -e 2 . 



In particular, these formulas hold when e x =U 1 (p), e 2 = U 2 (p), in which 
case e 3 = U x (p) X U 2 (p) = U 3 (p). 



8 M334 II. 1 

There is no reason to restrict the definition of cross product to tangent 
vectors. As for the dot product, we can define it for pairs of points of E 3 
also. If we put U! = (1, 0, 0), u 2 = (0, 1, 0), u 3 = (0, 0, 1) then for points p 
and q of E 3 we define the cross product of p and q to be 

u 1 u 2 u 3 

pX q= pi p 2 p 3 

qi q2 q3 

= (P2q3 -P3q2)«i +(p3qi -piqsK + (piq2 -P2qi)u 3 
= (P2qs -Psqa.psqi -Piq3>piq2 -P2qi)- 

Useful results about determinants and cross products are contained in the 
following exercises, which you should attempt now. 

1. Page 49, Exercise 4. 

2. Page 49, Exercise 5 (first part). 

3. Page 49, Exercise 6. 

Additional Text 

The result of the following exercise will be needed later. 
4. Page 49, Exercise 7. 



Summary 

Notation 

p-q 

vp-w p 

IIpII 

llvpll 

d (p> q) 

*A 

v p X w p 
pXq 



Page 42, Definition 1.1 

Page 44, Definition 1.3 

Page 43, line 9 

Page 44, line 12 

Page 43, Definition 1.2 

Page 43, line -4 

Page 47, line 1 

Page 47, Definition 1.7 

Text, page 8 



Definitions 

(i) Dot product of points p*q 

of tangent vectors vw 

(ii) Norm of a point ||p|| 

of a tangent vector ||v|| 

(iii) Euclidean distance d(p, q) 



Page 42, Definition 1.1 
Page 44, Definition 1.3 

Page 43, line 9 
Page 44, line 12 

Page 44, Definition 1.2 



M334 II. 1 

(iv) e- neighbourhood 9l ( 

(v) Open set 

(vi) Orthogonal vectors 

(vii) Unit vector 

(viii) Frame, or orthonormal basis 

(ix) Orthonormal expansion 

(x) Attitude matrix 

(xi) Orthogonal matrix 

(xii) Transpose of A, t A 

(xiii) Cross product of points pXq 

of tangent vectors v X w 

(xiv) Triple scalar product' u«v X w 



Page 43, line -4 
Page 43, line -3 
Page 44, line -7 
Page 44, line -6 
Page 44, Definition 1.4 
Page 45, line -1 
Page 46, Definition 1.6 
Page 46, line -2 
Page 47, line 1 

Text, page 8 

Page 47, Definition 1.7 

Page 48, line - 3 of text 



Results 



(i) 



lip + qll < llpll + llqll (the triangle inequality) and 
llapH = |a| llpll. 

(ii) |v-w| < ||v|| ||w|| (Schwarz inequality). 

(iii) Uj(p), U 2 (p), U 3 (p) constitute a frame at p. 

(iv) Orthonormal expansion. If e^ e 2 , e 3 is a frame 
at p and v€T p (E 3 ), then 

v = (ve 1 )e 1 + (v-e 2 )e 2 +(ve 3 )e 3 . 

(v) If e lt e 2 , e 3 is a frame, v = 2 ajej and 

w = Z bjej, then 

v-w = a^! + a 2 b 2 + a 3 b 3 . 

(vi) If A is an orthogonal matrix, *A = A -1 . 

(vii) The dot product is symmetric, bilinear, positive 
definite. 

(viii) The cross product is alternating, bilinear. 

(ix) ||v X w|| = (vv ww- (vw) 2 )2 = ||v|| ||w|| sin #. 

(x) v X w is orthogonal to v and to w. 

(xi) v X w = ||vj| ||w|| sin # n, where n is obtained by the 
right-hand rule. 

(xii) v X w # if and only if v and w are linearly inde- 
pendent. 

(xiii) u«v X w ¥= if and only if u, v and w are linearly 
independent. 

(xiv) Interchanging any two of u, v, w reverses the sign 
of u-v X w but does not change its absolute value. 

(xv) If ej, e 2 , e 3 is a frame, 

e l -e 2 X e 3 =± 1. 



Page 43, line 11 
Page 44, line 13 
Page 45, line 6 



Page 45, Theorem 1.5 

Page 46, line 7 
Page 47, line 5 

Page 43, lines 1-7 
Page 47, line -11 

Page 47, Lemma 1.8 and 
Page 48, line -9 

Page 47, Lemma 1.8 

Text, page 7 
Page 49, Exercise 5 
Page 49, Exercise 4(b) 
Page 49, Exercise 4(c) 
Page 49, Exercise 6 



10 

(xvi) If e 1} e 2 are orthogonal unit vectors and 

e i X e 2 = e 3 , then ej, e 2 , e 3 is a frame, and 



M334 II. 1 



ej X e 2 = e 3 ; e 2 X ei = ~e 3 ; 
e 2 X e 3 = ei; e 3 X e 2 = -e 1 ; 
e 3 ^ e i = e 2> Ci X e 3 = - e 2 . 



Text, page 7 



(xvii) If u is a unit vector, any vector v may be expressed 
uniquely as 

V = (V'U) u + v 2 , 

where v 2 is orthogonal to u. Page 49, Exercise 7 



Techniques 

(i) Evaluation of dot and cross products and norms. 

(ii) Evaluation of triple scalar products. 

(iii) Expansion in terms of an orthonormal basis. 

(iv) Determination of the attitude matrix and use of 
the result t A = A" 1 . 



Page 42, Definition 1.1 and 
Page 47, Definition 1.7 

Page 49, Exercise 4(a) 

Page 45, Theorem 1.5 

Page 46, Definition 1.6 



Exercises 

Technique (i) 

5. Page 48, Exercise 1. 



Technique (ii) 
6. 



If v and w are as in the preceding exercise and u = (-3, 1,-1) at the same 
point, compute u«v X w. 



Technique (iii) 

7. Page 49, Exercise 3. 



Technique (iv) 

8. Find the attitude matrix A of the frame in the preceding exercise. What is its 

inverse? 



M334 II. 1 
Solutions 



11 



1. Page 49, Exercise 4. 

(a) Suppose u, v, w are tangent vectors at the point p. 

u = U!Ui(p) + u 2 U 2 (p) + u 3 U 3 (p). 
v X w = cjU^p) + c 2 U 2 (p) + c 3 U 3 (p) (*) 

where the cj are found from 



v X w = 



iMp) u 2 ( P ) u 3 ( P ) 

Vi v 2 v 3 

Wj w 2 w 3 



U-V X W = CjUj + C2U2 + c 3 u 3 

which is (*) with Uj(p) replaced by uj for i = 1, 2, 3: thus replacing 
Uj(p) by u[ in the determinant gives u«v X w. 



(b) 
(c) 
(d) 



u«v X w # <=*• the rows of the determinant are linearly independent 
<==>• u, v and w are linearly independent. 

Again, this follows from the properties of determinants: if any two 
rows in a determinant are interchanged the whole determinant 
changes sign. 



u X v«w = w-u X v, 
= u«v X w, 



because the dot product is symmetric, 
by part (c). 



Page 49, Exercise 5. 

We know that l|v X w|| = ||v|| ||w|| sin #, where # is the angle between v and w. 

v X w = <==> ||v X w|| = 

||v|| = or ||w|| = or sin # = 

v = or w = or & is a multiple of it. 

If v = or w = then v and w are certainly linearly dependent. If # is a 
multiple of tt then 



w 



w 



# = 



# = IT 



v and w are collinear, and so linearly dependent. Conversely, if v and w are 
linearly dependent, either v = or w = or sin # = 0. 

Thus v X w t£ <=> v and w are linearly independent. 



12 
3. 



M334 II.l 



Page 49, Exercise 6. 



e 2 X e 3 is orthogonal to both e 2 and e 3 . e 2 and e 3 span a plane in T p (E ), 
so the tangent vectors orthogonal to both e 2 and e 3 form a line in T p (E 3 ). 
We know that e 2 is orthogonal to both e 2 and e 3 and so e x spans this line: 
thus e 2 X e 3 must be a multiple of e x . 

lle 2 Xe 3 || = ||e 2 ||||e 3 ||sin|=l, 



so 



Thus 



e 2 X e 3 = ±e v 



ei-e 2 X e 3 = ie^ej =±1. 



If A is an orthogonal matrix, then it is the attitude matrix of a frame e^ 
e 2 ,e 3 , where the coordinates of ej with respect to Ux(p), U 2 (p), U 3 (p) form 
the ith row of A for i = 1, 2, 3. By the first part of the question, 
e 1 -e 2 X e 3 = ± 1. However, by Exercise 4(a) on page 49, 



e r e 2 X e 3 



en e 12 e 13 

e 21 e 22 e 23 
e 31 e 32 e 33 



= det A 



so the determinant of A is ± 1. 

We could have reached this last result in another way, using the facts (see 
Unit M201 5, Determinants and Eigenvalues, Section 1) that if A and B are 
square matrices of the same size then 



and 



det (*A) = det (A) 



det (AB) = det(A) det(B), 



Now, if A is orthogonal, *AA - I 



so 



so 



so 



so 



det (*AA) = det(I) = 1 
det (*A) det(A) = 1 



det (A) det(A) = 1 



det(A) = ±l. 



Page 49, Exercise 7. 

Choose v x = (v-u)u, v 2 =v - v 2 . Then certainly v = Vj + v 2 and v x is the 
component of v in the u direction, by definition. Moreover, v - \ x is the 
only possible choice for v 2 in order that \ = \ x + v 2 , and so this expression is 
certainly unique. 



M334 II. 1 



13 




v 2 



It remains to prove just that v 1 -v 2 = 0. Now 
vj-vj =x l -(x- vO 

= (vu)u • (v - (v-u)u) 

= (vu)(u-v - (v«u)u-u) 

= (vu)(u«v - v«u), because u-u = 1, 

= 0, because u»v = v«u. 

Page 48, Exercise 1. 

(a) vw = l.(- 1) + 2.0 + (- 1).3 = - 1 -3 = -4. 

(b) U^P) U 2 (p) U 3 (p) 
v X w= 1 2 -1 

-10 3 

= (2.3-(-l).0,(-l).(-l)- 1.3,1.0- 2.(-l)) 
= (6,-2,2). 

(c) llv|l = (i a +2 a +(-i) a )*=V6; pr^- 
IMi = ((-i) 2 + o 2 + 3 2 )^=ViO; IR^- 1 ^^- 



(d) 



|v Xw|| = (6 2 +(-2) 2 +2 2 )=V44 = 2VH. 

Alternatively, we can use Lemma II. 1.8: 

||v X w|| 2 = (vv)(w-w) - (v-w) 2 = ||v|| 2 ||w|| 2 - (vw) 2 = 6.10-(-4) 2 = 44. 

(e) If # is the angle between v and w, 

v»w = |jv|| ||w|| COS# 

so -4 =^6/10 cos # 

-4 -4 -2 

so costfsygo-a^g-^g- 

-1 

-1 

3 

= -3(2.3- (-1).0) + 1((-1) 2 - 1.3)- 1(1.0- 2.(-l)) 
= -3.6 + 1. (-2)- 1.2 =-22. 

Alternatively, since v X w has already been calculated in Exercise 5, we could 
go straight to 

u-vX w = (-3, 1,-1). (6,-2,2) 

= -3.6 + l.(-2) + (-l).2 = -22. 





Ul 


u 2 


u 3 




-3 


1 


•v X w = 


Vl 


v 2 


v 3 


= 


1 


2 




Wj 


w 2 


w 3 




-1 






14 

7. 



M334.II.1 



Page 49, Exercise 3. 

e 1 .e 1 = |(l 2 +2 2 +l 2 ) = l 

e 2 -e 2 =-|((-2) 2 + Q 2 +2 2 ) = 1 

e 3 .e 3 =!(l 2 +(-l) 2 +l 2 ) = l 

ei ' e2= 7678 ( - 2 + + 2) = 

e 2*e 3 = -75-75- (-2 + + 2) = 



1 1 



(1-2 + 1) = 



e3 ' ei ~VV6 

This shows that e 1} e 2 , e 3 constitutes a frame. 
By Theorem II. 1.5, 

v = (v«c 1 )c 1 + (v-e 2 )e 2 + (ve 3 )e 3 . 
v. ei =-L (6 + 2 -i)= 7 



v-e-- 



V6 



(-12-2) = 



V6' 
-14 



v ' e3= 73( 6-1 - 1)= 73 ; 



14 4 

'e, +— — e. 



V -V6 e i"V8 C2 V 3 " 3 * 
We can check this by working out the right-hand side directly: we obtain 

7 (1,2,1) 14 (-2,0,2) 4 (1,-1,1) 
V6 V6 7 s " V 8 \73" V 3 

=-^(1,2, l)-^(-2,0,2)+|(l,-l, 1) 



~ C> Q' C + O '» ^» 9 + *' «' 3 



= (6,l,-l)=v. 

8. We simply take the coordinates of c u e 2 , e 3 and write them as the rows of 

A. Thus 



1 



V6 V 6 V 6 



A = 



-2 



V8 



_J_ ^1 —L 

V 3 V 3 V 3 



M334 II. 1 15 

Since A is an attitude matrix it is orthogonal, and so we know t A = A * : that 
is, A~ * may be obtained from A by interchanging corresponding rows and 
columns. 



This gives 



,-i - 



1 


-2 


1 


V6 


V« 


V3 


2 





-1 
V3 


1 


2 


1 


V6 


V8 


V3 



16 M334 II.2 

II.2 CURVES 

Introduction 

This section depends heavily on Sections 1.4 and II. 1 but not at all on the later 
sections of Chapter I. 

Here we take up the story of curves left in Section 1.4, and extend the definitions 
and results of Section II. 1 to vector fields defined on curves. We immediately 
introduce a way of differentiating these vector fields; in particular this gives us the 
acceleration of a curve: the velocity and acceleration turn out to be the same as 
those of a particle travelling along the curve, as given in Unit MST 282 2, 
Kinematics, Section 1.1. 

All of this is simply giving us enough technical apparatus to establish the Frenet 
formulas, which are a core feature of the course, in the next section. 

READ: Section II.2 (pages 51-55) . 

Comments 

(i) Pages 51-52: unit-speed reparametrization This particular type of 
reparametrization is very important. From now on, whenever O'Neill mentions a 
"unit-speed reparametrization" he means not just a reparametrization which 
happens to have unit speed but a reparametrization of the type given in the proof of 
Theorem 2.1. There is a useful symmetry in this situation. If j3 is a unit-speed 
reparametrization of a we have 

|3(s) = a(t(s)); 

in this particular case s and t are inverse functions, and so we also have 

j5(s(t))=a(t): 

that is, a is also a reparametrization of ]3. We shall often make use of this fact. 

Theorem 2.1 is important because it is often sufficient to know that a unit-speed 
reparametrization of a curve ol exists, without computing it explicitly. It is this 
theorem which allows many proofs to begin: "We may assume a has unit speed". 
This considerably simplifies many things, but you should always check that the 
remark is justified. 

(ii) Page 52: Definition 2.2 A vector field on a curve should be distinguished 
from a vector field on E 3 . It is true that if o: is a curve and V is a vector field on E 3 
then the function Y defined by 

Y(t)=V(a(t)) 

is a vector field on a (see the diagram opposite); however not every vector field on a 
is necessarily the restriction of a vector field on E 3 . For example, it may happen 
that a crosses itself, that is there are different values t x and t 2 such that 

a(ti)=<*(t 2 ) = p; 



M334 II.2 



17 




Y(t) = V(a(t)) 



if Y is a vector field on a it is not obliged to take the same values at t t and t 2 , but if 
V is a vector field on E 3 then it has a unique value at p and so 
V(a(t!)) = V(a(t 2 )) = V(p). A good example of this is provided by the 
figure-of-eight curve given by 

a(t) = (sint,^sin2t, 0) (tGR) 

which crosses itself at = a(0) = a(ir): at this point the vector field a on a 
does take different values, for a'(0) = (1, 1, 0) while a'{ir) = (-1, 1, 0). 




► x 



(Hi) Page 54: Lemma 2.3 It is important at this point to be quite clear about 

O'Neill's slightly unusual use of the words "constant", "straight line" and "parallel". 
A curve a is a constant if it is of the form 

a: ti ► c 

that is, the image of a in E 3 is a single point. A curve a is straight line if and only if 
it is of the form 

a: 1 1 ► p + tq 

for some points p and q with q ¥= 0. The image of a is certainly the line joining 
p and p + q. Intuitively we might think of the curves 

/3: 1 1 > p + 2tq 



18 M334 II.2 

and 

7: 1 1 »p + t 3 q 

as the same as a, because their images are also the straight line joining p and p + q, 
but it is important to realize that, under Definition 1.4.1, they are different curves 
from a. In fact j3 is the reparametrization of a by the function 

h: t 1 — >2t 
and 7 is the reparametrization of a by the function 

k: t 1 ► t 3 . 

Are j8 and 7 straight lines? Well j(3 is, because j(3 can be written 

0= ti >p + t(2q), 

but 7 cannot be written in the form 

7: t 1 >a + tb 

for any points a and b so 7 is not, technically speaking, a straight line, merely a 
reparametrization of one. 

Finally, "parallel" has the meaning given on page 6, that of having the same vector 
parts. This is a more restrictive definition than that usually used in elementary 
geometry. For example, if 



V = (1, 0, 0) (0> 0> Q) y w 



W = ( 1 '0.0)(1,1,0) 

x =(2,0,0) (1>1>0) 

• ► 

y =(-l,0,0) (1)1?0) v 

then v is parallel to w but not x or y, even though they all point along the same 
direction. For tangent vectors at the same point we say they are collinear if they 
point along the same direction, that is if one is a scalar multiple of the other. The 
vectors w, x and y are all collinear. 

(iv) Curves — a Convention From now on we adopt the convention used by 

Spivak that a function defined on a subset of E has the whole of E as its domain 
unless otherwise specified.. Thus in Exercise 1 on page 55 the curve 

cx(t) = (2t,t 2 ,-£) 

is taken to be the curve 

a: t 1 >(2t, t 2 , y) (tGR). 



M334 II.2 19 

Additional Text 

The following exercises introduce results which will be needed later. 
Attempt them now. 

1. Page 56, Exercise 7. 

2. Y and Z are vector fields on a curve a. Prove that 
(YX Z)' = Y'X Z + YX Z'. 

Supplementary Comment 

(i) Page 51: the "standard theorem of calculus" This is a combination of 

several results from Spivak, which you should be able to accept. 



Summary 

Notation 

v 
s 

Y 
a" 

Definitions 

(i) Speed v 

(ii) Arc length 

(iii) Arc-length function s 

(iv) Arc-length parametrization 

(v) Unit-speed reparametrization 

(vi) Orientation-preserving reparametrization 

(vii) Orientation-reversing reparametrization 

(viii) Vector field (Y) on a curve 

(ix) Euclidean coordinate functions of a vector field on 
a curve 

(x) Derivative of a vector field on a curve 

(xi) Acceleration (a ") of a curve 

(xii) Parallel vector field 

(xiii) Collinear tangent vectors 



Page 5 1 , line 4 
Page 5 1 , line - 7 
Page 52, Definition 2.2 
Page 54, line 7 



Page 5 1 , line 4 
Page 5 1 , line 1 1 
Page 5 1 , line - 7 
Page 52, line 7 
Text, page 16 
Page 52, line -12 
Page 52, line -11 
Page 52, Definition 2.2 

Page 53, line 9 
Page 54, line 3 
Page 54, line 7 
Page 54, line - 10 
Text, page 18 



20 
Results 

(i) 



(iii) 
(iv) 



Any regular _curve has a unit-speed repara- 
metrization. 

If (3 is a unit-speed reparametrization of a, then a is 
a reparametrization of jS. 

If a has unit speed, then s(t) = t. 

The linearity and Leibnizian properties of 
differentiation of vector fields on curves. 



(aY + bZ)' = aY' +bZ' 
(fY)' = f'Y + fY' 
(Y-Z)' =Y'.Z + Y-Z' 
(YXZ)' =Y'XZ + YX Z'. 
(v) a is constant if and only if a = 0. 

(vi) a is a straight line if and only if a =fc and a" = 0. 
(vii) llY II is constant if and only if Y« Y' = 0. 
(viii) Y is parallel if and only if Y' = 0. 
(ix) Y(h)' = h'Y'(h). 



M334 II.2 



Page 51, Theorem 2.1 



Text, page 16 



Page 54, lines 12-17 
Text, page 19 
Page 54, Lemma 2.3(1) 
Page 55, Lemma 2.3(2) 
Page 54, lines - 14 to - 12 
Page 55, Lemma 2.3(3) 
Page 56, Exercise 7 



Techniques 

(i) Calculation of arc-length. 

(ii) Finding an arc-length function s = s(t). 

(iii) Unit-speed reparametrization. 

(iv) Expression of a vector field on a curve in terms of 
its Euclidean coordinate functions. 

(v) Algebra of vector fields on a curve. 

(vi) Differentiation of vector fields on a curve 

(a) from first principles 

(b) using linearity and Leibnizian properties. 

(vii) Calculation of velocity, speed, acceleration. 



Page 5 1 , line 1 1 

Page 51, proof of Theorem 2.1 

Page 51, Theorem 2.1 

Page 53, line 7 

Page 53, last paragraph 



Page 54, line 3 

Page 54, lines 12-17 and 
Text, page 19 

Page 5 1 , line 4 and 
Page 54, line 7 



Exercises 

Technique (i) 

3. Page 55, Exercise 4. (Note that "log" means "natural logarithm".) 



Techniques (ii) and (iii) 

4. Page 55, Exercise 3. 



M334 II.2 21 

Techniques (iv) and (v) 

5. Page 55, Exercise 6. 



Techniques (vi)(a) and (vii) 

6. Page 55, Exercise 2. The required cone is the set of points in E 3 satisfying 
x 2 + y 2 = z 2 . 

Technique (vi)(b) 

7. Let Y and Z be the vector fields on the curve a(t) = (cos t, sin t, t) given by 
Y(t) = (cost, sint, l) a (t)' 

Z(t) = (sin t, t, cos t)a( t ). 

Find Y'(t) and Z'(t) and hence V(t) when V is each of the following: 

(a) V(t) = Y(t) + Z(t) 

(b) V(t) = t 2 Y(t) 

(c) V = Y-Z 

(d) V = Y X Z. 



Theory Exercises (omit if you are short of time). 

8. Y is a vector field on a curve a. Prove that if ||Y|| is constant and Y' is never 
zero then Y" is never orthogonal to Y. (HINT: Differentiate Y"Y = con- 
stant.) 

9. Page 56, Exercise 9. (HINT: For the "if" part of (b), let 7 be a unit-speed 
reparametrization of a and show that 7 is a straight line by showing that 7 
is both collinear and orthogonal to 7'. By result (ii) a is a reparametrization 
of 7.) 



Solutions 

1. Page 56, Exercise 7. 

Y(t) is a tangent vector at a(t), so Y(h(s)) is a tangent vector at a(h(s)), so 
Y(h) is a vector field on the curve a(h). 

We can write Y(t) = (yi(t), y 2 (t), y 3 (t)) a ( t ) 

so 

Y , (t) = (y' 1 (t),yi(t),y , 3 (t)) a ( t ). 



22 M334 II.2 

Now 

Y(h)ts) = ( yi (h(s)), y 2 (h(s)), y 3 (h(s))) a(h(s)) 



so 



Y(h)'(s) = (y'i(h(s)) h'(s), yi(h(s))h'(s), yg(h(s))h'(s)) a(h(s)) 



by the definition of 
differentiation of a 
vector field on a curve 
and by the chain rule 



= h '( s )(yi( h (s)), yi(Ms)), y3(h(s))) a (h( s )) 
= h'(s)Y'(h(s)) 



so 



Y(h)' = h'Y'(h). 
Note that this result is, not surprisingly, yet another form of the chain rule! 

2. If 

Y(t) = ( yi (t),y 2 (t),y 3 (t)) 
and 

Z(t) = ( Zl (t), z 2 (t), z 3 (t)) 
then 

(Y X Z) (t) = (y 2 (t)z 3 (t)- y 3 (t)z 2 (t), y 3 (t) Zl (t) - Yl (t)z 3 (t), 

yi(t)z 2 (t)-y 2 (t)z 1 (t)). 

To differentiate Y X Z we differentiate its coordinate functions. The first of 
these is 

y 2 z 3 -y 3 z 2 , 
whose derivative is 

y 2 z 3 + y 2 z 3 - y 3 z 2 - y 3 z 2 , by the Leibnizian rule for derivatives 

of functions from R to R, 

= (Y2Z3 ~ y'&l) + (Y2Z3 " Y3Z2)- 

The other two coordinate functions are similar, and so we obtain 

(Y X Z)'(t) = (y 2 (t)z 3 (t) - y 3 (t)z 2 (t), y 3 (t) Zl (t) - yi(t)z 3 (t) , 

y'i(t)z 2 (t)-y 2 (t) Zl (t)) 
+ (y 2 (t)z 3 (t) - y 3 (t)z 2 (t), y 3 (t)zi(t) - yi (t)z 3 (t), 

yi(t)z 2 (t)- y 2 (t)zl(t)) 
= Y'(t) X Z(t) + Y(t) X Z'(t) 

= (Y'XZ + YX Z')(t). 
Thus 

(YXZ)' = Y' X Z + YX Z'. 



M334 II.2 23 

3. Page 55, Exercise 4. 

a(t) = (2t,t 2 ,logt),t>0. 

cx(l)=(2, l,0)=p 

<x(2)=(4,4,log2) = q, 

so the curve passes through p and q. 

a'(t) = (2,2t,i) a(t) 

|la'(t)|| = (4 + 4t 2 + p )"* = 2t +i since t > 0. 

The arc-length between p and q is the arc-length from t = 1 to t = 2, which is 
"2 



i 



2t + i dt = 



t 2 +logt 



2 

J 1 



= 4 + log 2 - 1 



3+ log 2. 



4. Page 55, Exercise 3. 

<x(t) = (cosh t, sinh t, t) 
c/(t) = (sinh t, cosh t, 1) a /j-\ 

||a'(t)|| = (sinh 2 t + cosh 2 t + 1)3 = (2 cosh 2 t)i 
= >/2 cosh t. 



because 1 + sinh 2 t = cosh 2 t, 



s(t) = \ ||a' (u) || du = I J 2 cosh u du = J 2 sinh t 
Jo JO 



so 



t(s) = sinh * —t~. 



The unit-speed reparametrization of a. based at t = is 

0(s) = a(t(s)) = (cosh(sinh _1 ~tt), sinh(sinh _1 ~jz), sinh -1 -j^) 



(using the identity cosh 2 =1 + sinh 2 for 6 = sinh * ~J7r), 



24 M334 II.2. 

5. Page 55, Exercise 6. 

(a) Y(t) = - a(t) at a(t) 

= (- cos t, - sin t, - 1) a u\ 

= -cost U^aft)) - sint U 2 (a(t)) - t U 3 (a(t)) 
= - cos tU]- sin t U 2 - t U 3 
using the convention mentioned in the middle of page 53. 

(b) a'(t) = (-sint, cost, l) a ( t ) 
a"(t) = (-cos t, -sin t, 0) a u\ 

Y(t) = (cost - sintJUj + (cost + sint)U 2 + U 3 . 

(c) Any tangent vector at a(t) orthogonal to both a'(t) and a"(t) is a 
scalar multiple of a'(t) X a '(t), if this is not zero. 

a'(t) X <x"(t) = (sint, -cost, 1) a u\ 

||oc'(t) X a"(t)|| 2 = sin 2 t + cos 2 t +1 = 2 

so a unit vector in the direction of a'(t) X a"(t) is ± (ljy/2)oi'(t) x a " ( l )- 



Thus Y(t) = ± ( sin L U ' - cos / t U 2 + Ih. 
W \ V2 V2 V2 



(d) Y(t) = (a(t+ir)-a(t)) a(t) 

= ((cos(t + 7r), sin(t + 7r), t + 7r) - (cos t, sin t, t)) a ( t \ 
= ((-cos t, -sint, t + 7r) - (cos t, sin t, t)) a u\ 
= - 2cos t Ui - 2sin t U 2 + 7rU 3 . 

Page 55, Exercise 2. 

Let p = a(t) = (t cos t, t sin t, t). 

The usual reason for the presence of the sine and cosine functions in the 
Euclidean coordinate functions of a curve is to take advantage of the 
identity sin 2 + cos 2 = 1. Thus we see that 

p 2 + p 2 = t 2 (cos 2 1 + sin 2 1) = t 2 = p 2 , 

that is, every point p on the curve satisfies 

Pi + P^=P 2 3 - (*) 



M334 II.2 25 

If C is the set of points satisfying (*), C intersects the plane p 3 = r in the 
circle p 2 + p 2 = r 2 , for each real number r. Thus C is the circular cone with 
axis the z-axis, vertex the origin, and half-angle — . Every point of the curve 
lies on C; the vertex (0, 0, 0) is a(0). 

a'(t) = (cos t - t sin t, sin t + t cos t, 1) a ^ t ) 

a'(0) = (1,0,1) ( o,o,0) 

Ha'(0)|| = (l 2 +l 2 f =V2; 

a"(t) = (- 2sin t - t cos t, 2cos t - t sin t, 0) a / t \ 
a"(0) = (0,2,0) (0,0,0)- 



7. We first evaluate Y' and Z', differentiating each coordinate function 

Y(t) = (cost, sint, l)o;(t) 
Y'(t) = (-sint, cost, 0) a n\ 

Z(t) = (sint, t, cost) a ( t \ 
Z (t) = (cost, 1, -sint)^/^. 

(a) V(t) = Y(t) + Z(t) 

so V'(t) = Y'(t) + Z'(t), by linearity, 

= (cos t - sin t, cos t + 1 , - sin t) (x(t)- 

(b) V(t) = t 2 Y(t) 
so 

V'(t) = 2t Y(t) + t 2 Y'(t), by the Leibnizian property, 

= 2t (cost, sint, l)(Wt) + t 2 (-sint, cost, 0) a n\ 
= (2t cost- t 2 sint, 2t sint + t 2 cost, 2t) a uy 

(c) V(t) = Y(t)-Z(t) 

so 
V'(t) = Y'(t)-Z(t) + Y(t)-Z'(t), by the Leibnizian property, 

= (- sin t, cos t, 0) • (sin t, t, cos t) + (cos t, sin t, 1 ) • (cos t, 1 , - sin t) 

= - sin 2 1 + t cos t + + cos 2 1 + sin t - sin t 

= cos 2 1 - sin 2 1 + t cos t„ 



26 M334 il.2 

(d) V(t) = Y(t) X Z(t) 



so 



V'(t) = Y'(t) X Z(t) + Y(t) X Z'(t), by the Leibnizian property, 

= (-sint, cost, 0) a / t \ X (sint, t, cos t) a u\ 
+ (cost, sint, l)a( t ) X (cpst, 1, -sint) a ( t ) 
= (cos 2 1, sin t cos t, - 1 sin t - sin t cos t) a n\ + 

(- 1 - sin 2 1, cos t + sin t cos t, cos t - sin t cos t) a / t \ 
= (cos 2 1 - sin 2 1 - 1 , cos t + 2sin t cos t, cos t - t sin t -2sin t cos t) a / t \ 
= (- 2sin 2 1, cos t + sin 2t, cos t - tsin t - sin 2t) a n\. 



8. If ||Y|| has the constant value c, then 



Y-Y = c 2 . 



Differentiation gives 

Y'-Y + Y-Y' = 

so 

Y-Y' = 0. 

Further differentiation gives 

Y'-Y' + Y-Y" = 0. 



If Y(t) is orthogonal to Y"(t) then (Y'Y")(t) = 0. This forces (Y'"Y')(t) = 0, 
that is || Y (t)|| = 0. The norm of a vector is zero if and only if that vector is 
itself zero, so this implies Y'(t) = 0, which is contrary to assumption. Thus 
Y(t) cannot be orthogonal to Y"(t). 



Page 56, Exercise 9. 

(a) | \a || is a constant 



a -a is a constant 



(«'.<*')' = 



a -a + a. -a =0 



a -a =0 

a is always orthogonal to a . 



M334 II.2 



27 



(b) Suppose j3 is the straight line t>-> p + tq, and a is the reparametrization 0(h) 

of j3. Then 

a(u) = p + h(u)q 

a'(u) =h'(u)q a ( u ) 
a"(u)=h"(u)q«( u ). 

Then a'(u) and a"(u) are both scalar multiples of the tangent vector q a ( u ), 
and so they are collinear. Conversely, we wish to show that the collinearity 
of a and a" implies that a is a reparametrization of a straight line. By the 
remark about symmetry in Comment (i), it is enough to show that a has a 
unit-speed reparametrization 7 = a(h) which is a straight line. Consequently 
we choose 7 to be a unit-speed reparametrization of a: Theorem II.2.1 tells 
us that we can choose. such a 7. 

Now 7(s) - a(h(s)) 

7'(s) = h'(s) a'(h(s)), by Lemma 1.4.5, 

7"(s) = h"(s) a (h(s)) + (h'(s)) 2 a"(h(s)), by the Leibnizian property. 

a" (h(s)) and a'(h(s)) collinear 

=> a"(h(s)) is a scalar multiple of a'(h(s)) 

=> t"(s) is a scalar multiple of ot (h(s)) 

=> 7"(s) and 7'(s) are collinear. 

However, 7 has unit speed, that is \\y' \\ = 1, so part (a) tells us that 7"(s) is 
orthogonal to 7'(s). The only way that 7" (s) can be both collinear and 
orthogonal to 7'(s) is that 7"(s) = 0: hence by Lemma 11.2.3(2) 7 is a straight 
line. 



28 M334 II.3 

II. 3 THE FRENET FORMULAS 

Introduction 

This section carries straight on from Section II. 2. The Frenet formulas for unit-speed 
curves are the first really important result in the course. You should try to become 
thoroughly familiar with the concepts of tangent, principal normal, binormal, 
curvature K and torsion r, which are collectively called the Frenet apparatus, and be 
able to find them for any unit-speed curve. 

We use these ideas to define the Frenet approximation to a curve, which is a type of 
Taylor approximation. You have met Taylor approximations before in Unit Ml 00 
14, Sequences and Limits II or Unit MST 281 10, Taylor Approximation, and Unit 
M231 13, Taylor Approximations. The Frenet approximation is less important than 
the Frenet apparatus, except in so far as it helps one to visualize the shape of a curve 
near a point and suggests various characterizations of curves in terms of their 
curvature and torsion. Such a characterization is a statement of the form: 

j3 is a curve of a particular type (e.g. plane curve, cylindrical helix) if and 
only if the curvature (k) and torsion (r) of j3 satisfy a particular condition 
(e.g. r = 0; or t/k is constant). 

The fact that there are such characterizations indicates why k and r are important 
functions associated with a curve: we shall see just how important they are in 
Chapter III. The section concludes with some characterizations of curves: you should 
note the methods of proof as much as the results. Further characterizations follow 
in the exercises and in the next section. 

READ: Section II.3 (pages 56-63) . 

Comments 

(i) Pages 57-8: curvature and principal normal The definition of K as ||T'|| 

forces us to avoid points where k is zero, for at such a point not only is N undefined 
but K may not be differentiable (because the square-root function is not differenti- 
able at zero). Moreover, N may suddenly change from one side of the curve to the 
other near a point where k is zero, as shown in the diagram opposite. 

Consequently the footnote on page 57 is very important: we restrict our attention 
to portions of curves where the curvature is positive. 

It is possible to develop a theory where k is allowed to become zero' and even 
negative, in which N is defined for many curves even at points where K = 0. In this 
theory k and N are still differentiable at such points. O'Neill does not take this 
approach (except for curves in E 2 — see the optional section in the exercises), but if 
you are interested you can read about it in T. J. Willmore: An Introduction to 
Differential Geometry (O.U.P. 1964), Chapter I. 

(ii) Page 57: binormal We choose B to be T x N specifically to obtain a vector 

field which is orthogonal to T and N at each point. Since T(s) and N(s) are 
orthogonal unit vectors, Lemma II. 1. A tells us that B(s) is a unit vector orthogonal to 
T(s) and N(s), and moreover the following equations hold: 

TXN = B; NXT=-B; 
NXB = T; BXN = -T; 
BXT = N; TXB = -N. 



M334 II.3 



29 





Since T(s), N(s), B(s) form a frame at each point |3(s) we can extend the idea of 
orthonormal expansion by the pointwise principle to obtain the result that any vector 
field Y on has an orthonormal expansion of the form 

Y = (Y-T)T + (Y-N)N + (Y'B)B. 



(in) Page 59 onwards: the correspondence between points and tangent vectors 
The identification of the tangent vector v p with the point v, for working which is 
concerned just with the Euclidean coordinates, enables us to omit the subscripts to 
tangent vectors. This convention considerably simplifies manipulation and proofs. 
For example, O'Neill uses this convention throughout the proof of Corollary 3.5: 
without this convention the last line of page 6 1 would read 

^( s )^jS(s) = ns)-q j g(s) = 

and the second paragraph on page 62 would contain words to the effect that since B 
is parallel there is a fixed vector v such that B(s) = v^/ s \ and would then define f(s) 
as (j3(s) - j8(0))«v; this adds verbiage and symbols to the proof without enhancing our 
understanding of it. Consequently we use this convention whenever it is possible to 
do so without causing confusion. We note that by page 58 O'Neill is already writing 



tv \ _ i a. • s a s b 
T(s)-|- 7r sm-,-cos-,- j 



c' c 



rather than the technically more correct 



T(s)=--sin-, 



a s b 

- cos — , — 
c c' c 



s • s bs 
a cos — , a sin -, — 



In the last paragraph on page 59 we take this identification of tangent vectors with 
points a little further. We associate the tangent vector Y(s) = (y (s), y 2 (s), y3(s))j3( s ) 
with the point ( yi (s), y 2 (s), y 3 (s)) of E 3 . 



30 



M334 II.3 




We simply take all the tangent vectors Y(s) and move the point of application j3(s) to 
the origin; the new end points of the tangent vectors become the new curve. If we 
call this new curve a, then a is parametrized in such a way that a(s ) is the point 
with the same Euclidean coordinates as the tangent vector Y(s ). This technique 
allows us to transfer certain results about curves to vectors fields on curves, and vice 
versa. 



Additional Text 

The following exercise introduces a result that we shall need in the next section, so 
attempt the exercise now. Do not spend too long on it, and be sure to read the 
solution. 



Page 65, Exercise 7. 



Supplementary Comments 

(i) Page 57: line -2 The "orthonormal expansion" referred to is that of B' in 

terms of T, N, B, as mentioned in Comment (ii). 

B' = (B'-T)T + (B'-N)N + (B'-B)B 
= (B'-N)N if B'-T = B'*B = 0. 

(ii) Page 59: line 5 Looked at from above we have: 



„ „ >.~ cos -, a sin— 0) 

N(s) = (-cosl,-sin-l, 0)^ A c c 




The normal is in the opposite direction to the projection of the position vector |3(s) 
on the xy-plane, and so it always points towards the z-axis. 



M334 II.3 31 

(iii) Page 61: line 4 This is not the Taylor approximation to |8 of degree 3: that 

would be 

0(s) -0(0) + (s- Kl s 3 )T + (/c | + ^(0)s 3 )N +k t jB . 

We are interested in the approximation of the component of |8(s) - |3(0) in each of 
the directions T , N , B , so we keep just the term of lowest degree in each 
component. 



Summary 

Notation 

T 

K 

N 
B 
r 



Page 56, line -4 
Page 57, line 2 
Page 57, line 6 
Page 57, line 8 
Page 58, line 3 



Definitions 



(i) 


Unit tangent T 


(") 


Curvature k 


(iii) 


Principal normal N 


(iv) 


Binormal B 


(v) 


Frenet frame field T, N, B 


(vi) 


Torsion r 


(vii) 


Frenet approximation 


(viii) 


Tangent line 


(ix) 


Osculating plane 


(x) 


Frenet apparatus T, N, B, 


(xi) 


Plane curve 


(xii) 


Spherical curve 



Page 56, line -4 
Page 57, line 2 
Page 57, line 6 
Page 57, line 8 
Page 57, Lemma 3.1 
Page 58, line 3 
Page 61, line 5 
Page 61, line 10 
Page 61, line 14 

Page 61, line -12 
Page 63, last two lines 
of text 



Example 



(i) 



The unit-speed helix 



„/ \ i s • s bs 

j 3(s) = |acos_, asm-,— 



where c = a + b , has constant curvature a/c and 
constant torsion b/c 2 . 



Page 58, Example 3.3 



(i) The Frenet formulas 






T' 


= 


/cN 




N' 


= - kT 




+ tB 


B' 


= 


tN 





(ii) The Frenet approximation: for small s, 



Page 58, Theorem 3.2 



j8(s) - |S(0) + sT(0) + fc(0)| N(0) + k(0)t(0)| B(0). Page 61, line 4 

(iii) A unit-speed curve with positive curvature is a 

plane curve if and only if r = 0. Page 61, Corollary 3.5 



32 M334 II.3 

(iv) A unit-speed curve has zero torsion and constant Page 59, third 

curvature k if and only if it is part of a circle with paragraph, and 

radius l//c. Page 62, Lemma 3.6 

(v) A spherical unit-speed curve has curvature > 1/a, Page 63, last 

where a is the radius of the sphere. paragraph 

(vi) If a =o:(h) and a are both unit-speed curves, then T 
= eT(h), N = N(h), B = e'B(h), k = /c(h), r = r(h), 

where e = h' = ± 1. Page 65, Exercise 7 



(vii) T = NXB, N = BXT, B = TXN. Text, page 28 

(viii) If Y is a vector field on a, 

Y = (Y-T)T + (Y-N)N + (Y-B)B. Text, page 29 

Techniques 

(i) Finding the Frenet apparatus T, N, B, k, t, by dif- 

ferentiation, followed by taking norms and cross 
products. Page 58, Example 3.3 

(ii) Identification of a vector field on a curve with Page 59, last 
another curve. paragraph 

(iii) The following two techniques are useful in 
characterizations : 

(a) to show a function is constant, show that 
its derived function is zero, simplifying the 
derived function by using the Frenet 

formulas; Page 62, lines 6-11 

(b) prove that "type of curve" => "properties 
of k and r" first, as the proof should give 

you information which will indicate the Page 62, proof of 
method of proving the converse. Lemma II. 3. 6 



Exercises 

Technique (i) 

2. Page 63, Exercise 1. It is sufficient to show that the curve is part of a circle, 
and find its centre and radius. 

Technique (ii) 

3. For j3 as in the preceding exercise, let o be the curve whose coordinate 
functions are given by a(s) = T(s). Find the speed of a. 

Technique (iii) 

4. Page 65, Exercise 10. 



M334 II.3 33 

Optional Section — Curves in the Plane 

5. Curves in the plane are treated somewhat differently from those in E .If 

you would like to make a small sub-study of them, read the definitions given 
in Exercise 8 and then do that exercise. More exercises on curves in the plane 
will be provided later. Note that, because E 2 is two-dimensional, N spans the 
set of vectors orthogonal to T: T' is orthogonal to T because ||T|| = constant, 
and so T' = /cN for some k. 



Solutions 

1. Page 65, Exercise 7. 

(a) a = <x(h) 

so a' = h'a'(h), by Lemma 1.4.5, 

so ||a'(s)|| = |h'(s)i ||a'(h(s))|| for all s. 

a is a unit-speed curve, so ||a'(h(s))|| = 1 for all s; S is a unit-speed 
curve, so ||R'(s)|| = 1 for all s: thus |h'(s)| = 1 and so h'(s) = ± 1. Since 
h is differentiable and has an interval as its domain, h'(s) cannot be 
+1 for some values of s and - 1 for other values of s. So we can write 
h' = e, where e = ± 1 independent of s. Then h(s) = es + h(0) = ± s + s 
where s = h(0). 

(b) The salient fact from part (a) is that h'(s) = e. 
T(s) = a(h)'(s) = h'(s)a'(h(s)), by Lemma 1.4.5, 

= cT(h(s)). 

f '(s) = eT(h)'(s) = eh'(s)T'{h(s)), by Exercise IL'2.7, 

= e 2 T'(h(s)) = T'(h(s)): 

thus 

7c(s)N(s) = ic(h( S ))N(h(s)): (*) 

taking norms we find 

5c(s)i ||N(s)|| = |K(h(s))| HN(h(s))|| 

so |jc(s)| = k(h(s))| since ||N(s)|| = ||N(h(s))|| = 1: 

/c(h(s)) and k(s) are positive so we may remove the modulus signs, 
obtaining 

Jc(s) = K(h(s)). 

Now this equation and (*) together imply N(s) = N(h(s)) since 
/c(h(s)) is not zero. 

B(s) =T(s) XN(s)=eT(h(s)) X N(h(s)) = eB(h(s)). 

B'(s) = eB(h)'(s) = eh'(s) B'(h(s)), by Exercise II.2.7, 

= c 2 B'(h(s)) = B'(h(s))=-r(h(s))N(h(s)): 



34 



thus 



M334 IL3 



but 



so 



-?(s).N(s)=-T(h(s))N(h(s)), 



N(s) = N(h(s)) 



f(s) = r(h(s)). 

The importance of this result is that if a and a are two different 
unit-speed reparametrizations (in the sense of Theorem II.2.1) of 
some curve 7, then a. and a have the same Frenet apparatus at each 
point of the curve. 



3. 



Page 63, Exercise 1. 

4 3 

|3(s) = (t-cos s, 1 -sin s, -v-cos s). 

|3'(s) = (-- t sin s, -cos s,-^ sin s) 

H]3'( s ) II = (|f sin 2 s + cos 2 s + ^ sin 2 sf = (cos 2 s + sin 2 s) 2 " = 1 , 
so |3 is indeed a unit-speed curve and 

T(s) = 0'(s). 

T'(s) = (— r- cos s, sin s,-jt cos s) 

-\a q 1 

k(s) = ||T'(s)|| = (J| cos 2 s + sin 2 s +-^ cos 2 s)* 



25 



= (cos s + sin 2 s) 2 = 1 



so 



N(s) = T'(s). 










u 2 


U 2 


u 3 


B(s) = T(s) X N(s) = 


4 . 
-z sins 
5 


- COSS 


3 . 
7 sms 
5 




4 

"e COS S 


sins 


3 

•= coss 

5 



= —r- (cos 2 s + sin 2 s),-„k (sin s cos s - sin s cos s), — r- (sin 2 s + cos 2 s) 

1 5' u ' 5/' 
B'(s) = so t(s) = 0. 

Since r = and /c is constant, Lemma II.3.6 tells us that j3 is part of a 

circle of radius— = 1 The proof of Lemma II.3.6 showed that the 

centre of the circle is given by ]3(s) + (1/k(s)) N(s), which in this case 

gives 

4 .3 4 3 

(— cos s, 1 - sin s, --r- cos s) + (— r cos s, sin s, -r- cos s) 

= (0,1,0). 

a is the curve whose coordinates are the same as those of the vector field T, 
and so 

lla'(s)||-||T'(s)|| = K(s). 

We saw in the preceding exercise that /c(s) = 1 and so ||a'(s)|| = 1 for all s, 
that is, o has unit speed. 



M334 II.3 35 

4. Page 65, Exercise 10. 

(a) If a lies on the sphere of centre c and radius r, then 
lloe (s) - c|| = r, 
that is, 

(a(s)- c).(a(s)- c) = r 2 . 

Using coordinates only and ignoring points of application (following 
Comment (iii)), we differentiate to obtain 

(a(s) - c) • a'(s) = 0, 

that is, 

(a(s) - c) • T(s) = 0. 

By the orthonormal expansion of a(s) - c in terms of the basis T(s), N(s), 
B(s), 

a(s) - c = a(s)N(s) + b(s)B(s) 
where 

a(s) = (a(s) - c)-N(s), b(s) = (a(s) - c).B(s). 

Differentiating again: 

a'(s) = a'(s)N(s) + a(s) N'(s) + b'(s) B(s) + b(s)B'(s), 

but 

«'(s) = T(s) 

so that 

T(s) = a'(s)N(s) + a(s)(-/c(s)T(s) + r(s)B(s)) + b'(s)B(s) - b(s)r(s)N(s). 

Since T(s), N(s), B(s), form a basis, we can equate coefficients on each side: 
coefficients of T(s) give 

1 =-a(s)K(s) 

and so 

coefficients of N(s) give 

= a'(s) - b(s)T(s) 
so 

Ms) = ^a'(s) 
= -a(s)p'(s). 
Thus a- c = - pN- p'aB. 
Since N and B are orthogonal, 

Hc*-c|| 2 =(-p) 2 +(-p'a) 2 
so 

r 2 = p 2 + {p'af. 



36 M334 II.3 

(b) We put 7(s) = a(s) + p(s)N(s) + p'(s)a(s)B(s) as we know by part (a) 
that 7(s) is constant when a is a spherical curve. We aim to show that 
7(s) is a constant if p 2 + (p'o) 2 = r 2 . We can achieve this if we can 
show that 7' = 0. 

Now 

V = ot' + p'N + pN' + (p'a)'B + p'aB' 

= T + p'N + p(-/cT + tB) + (p»'B - p'arN 
= (1 - p/<)T + (p' - p'or)N + (pr + (p'a)')B 
= (pr + (p'a)')B since p/c = or = 1. 

Now we use our information that p 2 + (p'o) 2 = r 2 . Differentiation gives 

2pp' + 2p'a(p'a)' = 0. 
so 

p + a (p'a)' = 0, because p' # 0, 

so 

pr + or {p'a)' = 

so 

pr + (p'a)' = 0, because or = 1. 

Thus 

7' = (pr + (p'a)')B = and so 7 is constant. 
Putting 7(s) = c gives 

a(s) - c =- p(s)N(s) - p'(s)a(s)B(s) 
and so, as before, 

l|a-c|| 2 = (-p) 2 + (-p'a) 2 

= P 2 +(p'o) 2 
= r 2 by assumption 

so 

l|a-c||=r, 

that is, a lies on the sphere with centre c and radius r. 



Page 65, Exercise 8. 

(a) N = (-y',x'),so 

N' = (-y",x"). 
However, T' = (x", y") = kN = »c(-y', x') so 

x"=-k Y ' 

a 1 

y = Kx 

and so 

N' = (-/ex', -Ky') =~k{x, y') = -kT. 




37 



Since T has slope angle <£ and unit length 

T = cos <p Ux + sin </? U 2 
and so 

T' = -simp <p'Vi + cos«p <p'U 2 . 

7T« 



N 



has slope angle (<p + ■*-) and unit length, so 

N = cos(<p+|)U 1 + sin(<p + |)U 2 
= - sin^U! + cos<pU 2 . 

T' = <p'N. 



Thus 



However, 

T' = kN, and so k = $'. 



38 M334 II.4 



II.4 ARBITRARY-SPEED CURVES 



Introduction 

This section extends the results of Section II. 3 to arbitrary speed curves. We need to 
use the chain rule for reparametrizations of curves (Lemma 1.4.5) and its analogue 
for vector fields on reparametrized curves (Exercise II.2.7) to adapt the Frenet 
apparatus to an arbitrary speed curve. We are searching for an effective procedure 
for computing the Frenet apparatus of an arbitrary regular curve, and, although we 
succeed in this, we lose the simplicity of the earlier Frenet formulas, with the result 
that the theory becomes cumbersome and the computations more involved and so 
liable to error. Although this whole section is important and very practical, if you 
find yourself becoming bogged down in calculus and computation give it a rest for a 
while and return later: the results of this section are not needed again until Section 
III.5. 

READ: Section II.4 (pages 66-74), omitting the last full paragraph on page 67. 



Comments 

(i) Page 66: the definitions The curvature, torsion etc. of a are those at the same 

point of the corresponding unit-speed curve a", that is, if p is the point <x(t) =o?(s(t)) 
on the curve, then the curvature at p is K(t) = lc(s(t)). There is a possible difficulty 
here, in that if we take different unit-speed reparametrizations of a. they may give a 
different Frenet apparatus. However, the result of Exercise II. 3. 7 tells us that so 
long as we restrict ourselves to unit-speed reparametrizations in the sense of 
Theorem II. 2.1 the Frenet apparatus is the same. Thus "the Frenet apparatus of a is 
the Frenet apparatus of any unit-speed reparametrization of a" is a valid definition. 

(it) Page 73: the orthonormal expansion ofu By Theorem II. 1.5 

u = (u-T)T + (u-N)N + (u-B)B 
= cos # T + aB for some a, because u*T = cos # and u«N = 0. 
Now u is a unit vector, so 

l = ||u|| 2 = cos 2 # + a 2 



so 



so 



a 2 = 1 - cos 2 # = sin 2 # 



a = ± sin #. 

If a = + sin # we have the required expression; if a = - sin # then, because cos (- #) = 
cos # and sin (-#)=- sin #, we can replace # by -& and obtain 

u = cos(-#)T + sin(-#)B. 

This is possible because we are allowing # to have any value. The recognition that we 
can just write sin # for a in such a situation should become second nature to you. 



M334 II.4 
Supplementary Comments 



39 



(i) Page 67: second paragraph You are not seriously expected to 

reparametrize the curve a(t) = (t, t 2 , t 3 ). If you attempted it and got stuck 
then you will understand O'Neill's point — there may not be explicit 
formulas for a. 

(ii) Page 69: the formula for K Since v = ||a'||, the equation 



(iii) 



yields 



la X a"|| = kv 3 



la X a 



la X dl 



a 



Ml 3 



Page 71: the example 

i 


i z 






^^^x ^ 


I y " 


/ \ b 
/ \ c 

"J 








**x 





Summary 

Notation 
a 

Definitions 

(i) Frenet apparatus for an arbitrary-speed curve 

(ii) Spherical image o 

(iii) Cylindrical helix 

(iv) Circular helix 

Results 

(i) T' = /cvN 

N'=-kvT + tvB 
B' = - rvN 



Page 71, line 10 



Page 66, first paragraph 
Page 71, line 10 
Page 72, Definition 4.5 
Text: page 41, Exercise 5 



Page 67, Lemma 4.1 



40 
(") 

(iii) 



(iv) 



a =-t-T + kv 2 N 



M334 II.4 



Page 68, Lemma 4.2 



T = 



B = 



a' X a' 



a' X a 



\a X a 
N = B X T 



a 



r = 



'||3 



(?L2L£Lk°L 
lla' X a"ll 2 



For a regular curve: 

K = if and only 

t = if and only 

K constant >0, r = if and only 

K constant >0, r constant =£0 if and only 

t/k constant if and only 



Techniques 

(i) Finding the Frenet apparatus of an arbitrary 

regular curve. 

(ii) Finding the spherical image of a curve. 

(iii) Finding the Frenet apparatus for a curve derived 
from some other curve in terms of the latter's 
Frenet apparatus: this involves repeated 
application of the Frenet formulas and taking 
higher derivatives. 

(iv) Finding u and # for a cylindrical helix by putting 
cot & = t/k and u = cos # T + sin # B. 



Page 69, Theorem 4.3 

if a is a straight line 

if a is planar 

if a. is part of a circle 

if o: is a circular helix 

if a is a cylindrical helix. 
Page 74 



Page 69, Theorem 4.3 
Page 71, second paragraph 



Page 71, last paragraph 



Page 73, Theorem 4.6 



Exercises 

Technique (i) 

1. Page 74, Exercise 2. This continues the example of Exercise II.2.3, whose 
solution is on Text page 23. 

Techniques (ii) and (iii) 

2. Page 75, Exercise 11. K a has been found on page 72, so all you need find is 
t . Make use of the working already done on pages 71-72. 

Technique (iii) 

3. Page 74, Exercise 8. 

Technique (iv) 

4. Page 75, Exercise 9(c). 



M334 II.4 41 

Theory Exercises (omit if you are short of time) 

5. Page 75, Exercise 10. This continues from Exercise II.4.8: a cylindrical helix 
is a circular helix if its cross-section curve y is part of a circle. 

6. Page 75, Exercise 12(a). This continues from Exercise II.4.11. 



Optional Section — Curves in the Plane 

If you have time to follow this optional line of study, you should also do: 

7. Let a be a plane curve, a a unit-speed reparametrization of a, that is a = a(s) 

where s is an arc-length function for a. Define 

T = T(s) 

K = k(s) 

N as in Exercise 8 on page 65. 
Prove that (a) a' = vT 

(b) T' = vkN 
(c)N' = -kvT. 

8. Page 75, Exercise 14. 



Page 75, Exercise 15. The central curve is defined in Exercise II.4.13. After 
you have sketched both curves, how do you account for the cusp (sharp 
point) in a* at a* (0)? 



Solutions 

1. Page 74, Exercise 2. 

a(t) = (cosh t, sinh t, t). 

We already know that: 

a'(t) = (sinht, cosh t, 1) 

v(t) = ^2 cosht 

s(t) = N /2sinht. 

Now 

a"(t) = (cosh t, sinh t, 0) 

a'"(t) = (sinht, cosht, 0) 

so a'(t) X a"(t) = (-sinh t, cosh t, sinh 2 1 -cosh 2 1) = (-sinh t, cosh t, - 1), 

||oc'(t) X a"(t)|| = (sinh 2 t + cosh 2 t + 1)2 =yj2 cosht, 

c/(t) X a"(t).a'"(t) = -sinh 2 t + cosh 2 t = 1. 



42 M334 II.4 

Now we simply apply Theorem II.4.3: 

T(t) = W^\ = 72 (tanh tj h sech t} (writing tanh for ^ and sech for ^ 

B W " ||a'(tj X a"(t)|| -J2 ( " tanht ' ^~ secht ) 

N(t) = B(t) X T(t) = |(2secht, 0, -2tanht) = (secht, 0, -tanht) 

,., n _ ll«'(t)Xa"(t)ll _ V2cosht _ 1 
{ } Ha'(t)|| 3 (V2 cosht)* 2cosh 2 t 

T W " ||a'(t) X a"(t)]| 2 " (V2 cosh t) 2 2 cosh 2 t 

Since we know s(t) =y/2 sinht, 2cosh 2 t = 2 + 2sinh 2 t = 2 + s 2 , 

soic(s) = T(s) = ^p. 



Page 75, Exercise 11. 

Let j3 have Frenet apparatus T, N, B, K, r. 

Making the usual identification of coordinates and ignoring both the 
distinction between points and tangent vectors and the points of application 
of tangent vectors, we can write, as on pages 71-72: 

a = T 

a = fcN 

a" = -k 2 T + k'N + ktB 

a'x o" = k 2 (kB + tT) 



\<j' xo"\\ = K 2 (K 2 + T 2 )\ 



Now 



so 



a'" = - 2kk'T - k 2 T' + k"N + k'N' + k'tB + kt'B + ktB' 

= - 2kk'T - k 3 N + k"N - k'kT + k'tB + k'tB + kt'B - kt 2 N 
= - 3/c/c'T + (k"~ k 3 - kt 2 )N + (kt + 2k't)B 

a' X a" -a'" = k 2 t(- 3kk') + k 3 (kt' + 2k't) 

= k 3 (kt' - tk) 

= k 5 (tIk)\ 

_ o'Xo"'o" , _ k 5 (t!k)' _ (t/k)' 
T ° lia'Xa"!! 2 k*(k 2 +t 2 ) k(1 + (t/k) 2 ) 

3. Page 74, Exercise 8. 

(a) We have to show that f(t)' = (7(t) - a(0))-u = 0. 

We know that 

f(0) = (7(0) - a(0))-u = s(0) cos # u-u = 



M334 II.4 43 

because s(0) = 0, and so it is sufficient to show that f is a constant 
function, i.e. that f = 0. 

f(t) = ( 7 (t)-a(0)).u 

= (a(t) - s(t)cos du- a(0))-u 

= (a(t) - a(0))-u - s(t)cos # because u-u = 1 ; 

f'(t) =a'(t).u- s'(t)costf 

= v(t)T(t)-u- v(t)cos# 

= v(t)cos & - v(t)cos # 

= 0. 

Hence 7(t) - a(0) is always perpendicular to u, that is, y lies in the plane 
through a(Q) orthogonal to u. 




(b) Let us suppose that a has unit-speed, so that s(t) = t and v(t) = 1. 

Using bars to denote the Frenet apparatus of 7: 

7(t) = a(t) - tcos # u 

7'(t) = a'(t) - cos # u = T(t) - cos # u. (*) 

v=H7'll 
* ((T - cos & u).(T - cos # u))2 
= (T-T - 2 cos & T-u + cos 2 & u-u)2 
= (1- 2 cos 2 »? + cos 2 #p 

= (1- COS 2 #)2 

1 

= (sin 2 ^)2" = |sin^|. 

Since v is constant, Lemma II.4.2 gives: 
7 "(t) = K(t)(v(t)) 2 N(t) 
= K(t) sin 2 # N(t), 



44 M334 II.4 

but our first expression, (*), for 7'(t) gives: 

T "(t) = T'(t) = K(t)N(t), 

so£(t)sin 2 #N(t) = /c(t)N(t). 

Taking norms gives K(t) = (K(t)/sin 2 #) since ic(t),7c(t), and sin 2 # are 

all positive. 



4. Page 75, Exercise 9(c). 

We use the technique used in the proof of Theorem II.4.6. We already have 
all the necessary information about a (see solution to Exercise II.4.2). 

r(t) = /c(t) = -j-, so r(t)/fc(t) = 1: we choose 

7r r(t) 

# =- so that cot# = 1 = -— x . 

4 K(t) 

Now we put U(t) = cos - T(t) + sin - B(t) 

= _1.J_ (tanht, l,secht)+— — (-tanht, 1,-secht) 

V2 V2 V2 V2 

= (0,1,0). 

This vector field U is parallel, as expected, so we put u = (0, 1, 0), and u is a 
fixed unit vector such that T(t)-u = cos (7r/4). 

7T 

7 (t) =a(t)- s(t)cos-u 

= (cosht, sinht.t)-^ sinht. -— (0, 1, 0) 

= (cosht, 0, t). 
7(t) - a(0) = (cosh t, 0, t) - (1, 0, 0) = (cosh t - 1, 0, t) 
so(7(t) -a(0)).u = (cosht- 1, 0, t)-(0, 1, 0) = 0, as required. 



5. Page 75, Exercise 10. 

Since K > and T # are constants, we can apply Theorem II.4.6: thus /3 is 
certainly a cylindrical helix with 

# = cof 1 (j\k\ u = cost? T(t) + sintf B(t). 

All that remains to show is that the cross-section curve 7 of is part of a 
circle. By Exercise 11.4.8(a), 7 is planar; by Exercise 11.4.8(b) the curvature 
of 7 is (K/sin 2 #), which is a constant: hence, by Lemma II.3.6 (which is 
unaltered if the unit-speed restriction is lifted), 7 is part of a circle. 



Page 75, Exercise 12(a). 

We may suppose the curve |8 has unit-speed. Let a be its spherical image. 

|8 is a cylindrical helix <=> t/k is constant (by Theorem II.4.6) 

<=> k is constant and r ff is zero (by Exercise II.4.11) 

<=» a is part of a circle (by Lemma II.3.6 and its converse). 



M334 II.4 45 

7. (a) a = 5(s) 



so 



a. = s a (s), by Lemma L4. 5, 

= vd'(s), because s' = v, 

= vT(s) 
= vT. 



(b) T = T(s) 



so 

-if tmt 



T' = s'T'(s), by Exercise II.2.7, 
= vT'(s) 

= v/c(s)N(s), because T' = kN, 

= vkN. 



(c) N = N(s) 



so 

r' 'tvt' 



N' = s'N'(s) , by Exercise II.2.7, 

= vfl'(s) 

= v(-Jc(s)T(s)), by Exercise II.3.8, 

= -v/cT. 



8. Page 75, Exercise 14. 

c/ = vT 
a" = v'T + vT' 

= v'T + v 2 kN 
Thusa".N = v 2 /c. 

However, in the plane N = J(T) = — , so 

v 






- ,,2 



_ a".J(c*') 



/c = 



v 3 



In terms of the coordinate functions, 

«' = ( x '> y')« 
](«') =(-y',x') a i 
v = (x' 2 +y' 2 j5 

a - ( x » y )« 

x y - x y 

so /c = , 

(x' 2 +y' 2 )t 



46 
9. 



M334 II.4 



Page 75, Exercise 15. 



(a) 






VK K 

= vT - - T - - N 

K K l 



K 

= -- 2 N. 



The tangent line to a* at a*(t) is the line through a*(t) in the direction 
a*'(t); the normal line to a at a(t) is the line through a(t) in the direction 
N(t), that is in the direction orthogonal to a' (t). Consequently these two 
lines are the same if and only if 

(i) a*'(t) and o/(t) are orthongonal: a*'(t)-a'(t) = 

(ii) there is a number u(t) such that a*(t) = a(t) + u(t)N(t). 




«'(t) 



a*'(t) 



a*(t) 



Firstly, let us check that a* does satisfy these conditions. 

a *'(t).a^t)="-^ 2 N(t).a'(t) = 
so (i) is satisfied. Moreover 



a*(t) = a(t) + — N(t) 

K(t) 



I 



so condition (ii) is satisfied with u(t) = —- . 

Conversely, suppose )3 is a curve satisfying (i) and (ii), that 



is 



and 



ff.a' = 



= a + uN. 



M334 II.4 47 

Then 

P' = ol' + u'N + uN' 
= vT + u'N - uv/cT 

= v(l - uk)T + u'N 



so 



ff-d = (v(l - u/c)T + u'N).(vT) 

= v 2 (l- UK). 

j3'«o:' = and v is non-zero, so this implies K = — , that is 

u 
1 

j3 = a + — N = a*. 



T(a') a"-T(a') 
(b) We have already shown that N = -±-J , K = — ^— ', 



v v~ 



1 
so a* = a + — N 
K 



v J J(a') 

= a + JV ' 



a -JK) v 

a' •a' 
a .J(a) 



(c) a(t) = (t + sin t, 1 + cos t) -TT<t<ir 

a'(t) = (1 +cost, -sint) 
a"(t) = (-sint, -cost) 

c/(t)-a'(t) = 1 + 2 cost + cos 2 t + sin 2 t = 2 + 2 cost 
J(a'(t)) = (sin t, 1 + cos t) 
a"(t)-J(a'(t)) = - sin 2 1 - cos t - cos 2 1 = - 1 - cos t. 

*/ v / v a '(t)-a'(t) 
Now a*(t) = a(t) + -n-^ — — ](a'(t)) 
U «"(t)'J(«'(t)) Jl W ' 

= a(t)-2J(cx'(t)) 

= (t + sin t, 1 + cos t) - 2(sin t, 1 + cos t) 

= (t - sin t, - 1 - cos t). 



48 



M334 II.4 




The cusp at a*(0) suggests that a* is not regular at that point. In fact 
that is the case, for 

o:*'(t) = (1 - cost, sint) 



so 



a*'(0) = (0, 0). 



M334 II.5 49 

II.5 COVARIANT DERIVATIVES 

Introduction 

This section depends on Sections 1.3 and II. 2; you also need to be familiar with the 
differential of a function from E 3 to R (Section 1.5). We introduce a new derivative, 
the covariant derivative. This is a very important idea in differential geometry, and 
serves as the starting point for a lot of more advanced work. The definition is similar 
to the other definitions of derivatives in the book, in that it is given in terms of a 

derivative we already know along the particular curve ti >p + tv. It generalizes the 

derivative of a vector field on a curve. 

READ: Section II. 5 (pages 77-80) . 

Comment 

(i) Page 78: the symbol V This is usually pronounced "del" or "nabla". 

Additional Text 

(i) Covariant Differential If W = SwjUi is a vector field on E 3 , the covariant 

differential, VW, of W is the function on tangent vectors defined by 

VW(v) = V V W. 

By Lemma II.5.2 V V W = Zv[wj] U^p) and so 

VWM^ZvhrJUifo) 

= 2dwi(v)Ui(p), by definition of dwj, 

= (ZdwiUi)(v). 

Thus we may write 

VW = SdwiUi- 

(n) A Useful Result The result of the following exercise will be used often 

later in the course. Attempt the exercise now. 

1. Page 81, Exercise 6(a). (HINT: Use Lemma 1.4.6.) 

Summary 

Notation 

V v w Page 78, line 2 

V V W Page 79, line 15 

V ™ Text, page 49 



50 
Definitions 

(i) Covariant derivative with respect to a tangent 

vector V V W 



M334 II.5 



Page 77, Definition 5.1 



(ii) Covariant derivative with respect to a vector field 

VyW Page 79, line 12 



(iii) Covariant differential VW 



Text, page 49 



Results 
(i) 



(i") 
(iv) 



If W = 2 wjUj, V V W = 2 v[wj] Ui(p) 
V v W = 2V[w i ]U i . 

Linearity and Leibnizian properties of the 
covariant derivative. 

With respect to a tangent vector v: 

V a v + bwY = aV v Y + bV w Y 

V v (aY + bZ) = aV v Y + bV v Z 

V v (fY)=v[f]Y( P ) + f(p)V v Y 

v[Y'Z] = V v Y'Z(p) + Y(p)'V v Z 

With respect to a vector field V: 

V fV + gW Y = fV vY + gVwY 
V V (aY + bZ) = aV V Y + bV V Z 
V v (fY) = V[f]Y + rV V Y 
V[Y-Z] =V V Y-Z + Y-V V Z 

If W = 2 wjUj, VW = 2 dwiUj. 

V(t)W = (w(«))'(t). 



Page 78, Lemma 5.2 
Page 79, line 18 



Page 78, Theorem 5.3 



Page 80, Corollary 5.4 

Text, page 49 

Page 81, Exercise 6(a) 



Techniques 

(i) Finding the covariant derivative with respect to a 

tangent vector or vector field, 



(a) 
(b) 

(c) 



(d) 



from first principles; Page 77, Definition 5.1 

using result (i) and calculating the 

directional derivatives from first principles; Page 78, Lemma 5.2 

using result (i), the linearity and Leibnizian 
properties of both the directional and 
covariant derivative, and the results 
Ui[f] =(3f/3xi),v[f] =Svi(8f/axi); 

by means of the covariant differential. Text, page 49 



Page 78, Lemma 5.2, 
Page 78, Theorem 5.3 and 
Page 80, Corollary 5.4 



M334 II.5 
Exercises 

Technique (i)(a) 

2. Page 80, Exercise 1(a). Work directly from Definition II.5.1. 



51 



Technique (i)(b) 



Page 80, Exercise 1(a). Use the result of Lemma II.5.2 but calculate the 
directional derivatives from first principles. 



Technique (i)(c) 

4. Page 80, Exercise 1(a). Now use all the results available to simplify the work! 

5. Page 80, Exercise 2(a). 

6. Page 80, Exercise 2(c). 

Technique (i)(d) 

7. Page 80, Exercise 5. 



Solutions 



Page 81, Exercise 6(a). 

LetW = 2w i U i . 

Then V (t) W = ?a'(t) [wj] Ui(a(t)), 

1 d(w:(a)) 
i dt 

= (?^(a(t))Ui( a (t)))', 
i 

= (W a )'(t). 



by Lemma II. 5. 2, 
by Lemma 1.4.6, 

by the definition of differentia- 
tion of a vector field on a curve, 



Page 80, Exercise 1(a). 
p = (l,3,-l),v = (l,-l,2),so 
p + tv = (l+t, 3- t,-l + 2t). 

W = x 2 U!+yU 2 

so W(p + tv) = (1 + t) 2 Uj (p + tv) + (3 -t)U 2 (p + tv). 
V v W = W(p + tv)'(0) 

= [2(l+t)U 1 (p + tv)-U 2 ( P + tv)]| t = 
= 2U,(p)-U 2 (p) 

= (2,-l,0) p . 



52 M334 II.5 

3. Page 80, Exercise 1(a). 

W = x 2 U! + yU 2 
so 

V V W = v[x 2 ] U^p) + v[y] U 2 (p) by Lemma II.5.2. 

v[x 2 ] =-x 2 (p + tv)| t = 
at 

-J t d + t)1 t .o 

= 2(l+t)| t = 
= 2; 

v[y]'= T-y(p + tv )|t = o 

at 
-f t (3-t)| t , 

= -i|t = o 

= -1. 
Thus 

V V W = 2U 1 ( P ) - U 2 (p) 
= (2,-l,0) p . 



Page 80, Exercise 1(a). 

V v W = v[x 2 ]U 1 (p)+v[y]U 2 (p). 

3(x 2 ) 3(x 2 ) 3(x 2 ) 

V X 2 =V!-r — (P)+V 2 — (p) + V 3 — — (p) 

ox dy dz 

= Vl 2x(p) 
=1.2.1 = 2; 

3y dy dy 

v [y] = v i r~(p) + v 2 r-(p) + v 3 — (p) 
ox oy dz 

= v 2 
= -1, 



so 



V v W = 2U 1 (p)-U 2 (p) 
= (2,-l,0) p . 



M334 II.5 53 

5. Page 80, Exercise 2(a). 

W = cos xUj + sin x U 2 
so 

V V W = V[cos x] U 2 + V[sin x] U 2 by Lemma II.5.2. 

V[cos x] = (-yUi + xU 3 ) [cos x] 

= -yU 1 [cosx] +xU 3 [cosx], by linearity, 
9(cos x) 9(cos x) 

= -y + x 

OX 8z 

= y sinx + 0; 

V[sin x] = (-yUi + xU 3 )[sinx] 

= -yUJsin x] + xU 3 [sin x] , by linearity, 

9(sin x) 3(sin x) 

= -y 1- x 

3x 3z 

= -y cos x + 0. 
Thus 

VyW = y sin x U l - y cos x U 2 . 

6. Page 80, Exercise 2(c). 

V v (z 2 W) = V[z 2 ] W + z 2 V V W by Corollary II.5.4. 

V[z 2 ] =2zV[z], by Corollary 1.3.4, 

= 2z(-yU 1 + xU 3 )[z] 

= -2yz U x [z] + 2xz U 3 [z] , by linearity, 

= -2yz.0 + 2xz.l 

= 2xz, 
so 

V v (z 2 W) = 2xz W + z 2 V v W 

= 2xz (cos x Ui + sin x U 2 ) + z 2 (y sin xUj-y cos x U 2 ) 

= (2xz cosx + yz^inxjUj + (2xz sin x - yz 2 cos x)U 2 . 

7. Page 80, Exercise 5. 

W = xy 3 U! - x 2 z 2 U 3 
so 

VW = d(xy 3 )U 1 - d(x 2 z 2 )U 3 

= (y 3 dx + 3xy 2 dy)U! - (2xz 2 dx + 2x 2 z dz)U 3 . 
(a) (y 3 dx + 3xy 2 dy) ((1, 0, -3)(_ 1} 2 , -1)) = 2 3 .1 + 3.(-l).2 2 .0 = 8 

(2xz 2 dx + 2x 2 zdz) ((1, 0, -3) ( _ ^ 2> _ 1} ) = 2.(- 1).(- 1) 2 .1 + 2.(- 1) 2 .(- l).(-3) = 4 
and so 

VW ((1, 0, -3) ( _ 1? 2 , -l)) = SU^p) - 4U 3 (p) 

= (8,0,-4) p . 



54 M334 II.5 

(b) (y 3 dx + 3xy 2 dy) ((-1, 2, -l) (lf 3j 2 )) = 3 3 .(-l) + 3.1.3 2 .2 = 27 

(2xz 2 dx + 2x 2 z dz) ((-1, 2, -1) (1> 3, 2 )) = 2.1.2 2 (-1) + 2.1 2 .2.(-1) = -12 

and so 

VW((-1, 2, -l)(i, 3, 2)) = 27U 1 ( P ) + 12U 3 (p) 

= (27, 0, 12) p . 



M334 II.6 55 

II.6 FRAME FIELDS 

Introduction 

This section does not depend on the previous two, but on Sections II. 1 and II.3. The 
idea of a general frame field (as opposed to the Frenet frame field of Section II.3) is 
formulated. 

READ: Section II.6 (pages 81-84) . 

Additional Text 

(i) The Pointwise Principle In this section O'Neill has used the pointwise 

principle to extend some familiar definitions and theorems from the set of 
all tangent vectors to E 3 to the set S of all vector fields V in E 3 . We recall 
that S is only a three-dimensional module (see Comment (ii) to Section 1.2 
(Text, page 10)) and so S is not isomorphic to E 3 . However, because each 
element of S can be expressed as 2f^U-, many of the ideas do carry over to S 
via the pointwise principle with no trouble. Some examples are given below. 

(a) The vector fields E u E 2 , E 3 form a basis for S if the tangent vectors 
E i (P)> E 2(p), Ea(p) form a basis for T (E 3 ) for each point p of E 3 . 

(b) The dot and cross products and norm are defined pointwise. That is, 
if V and W are vector fields then V«W is the function from E 3 to R 
defined by 

(V.W)(p) = V(p).W(p), 
V X W is the vector field defined by 

(VX W)(p)=V(p)X W(p), 
and ||V|| is the function from E 3 to R defined by 

IIVII(p) = HV(p)||. 

(c) Vector fields V and W are said to be orthogonal if V(p)-W(p) = for 
all p. For example, V X W is orthogonal to V and W. 

(d) Similarly the vector fields E t , E 2 , E 3 are orthonormal if 

that is (E i -E i )(p) = 1 for all p and for i =■ 1, 2, 3 and Ej, Ej are 
orthogonal for i ¥= j. 

(e) (Lemma 6.3). If Ej , E 2 , E 3 are orthonormal vector fields, any vector 
field V has an orthonormal expansion of the form 

V= (V.E,)^ + (V-E 2 )E 2 + (V.E 3 )E 3 

= f 1 E 1 + f 2 E 2 + f 3 E 3 
where fj = V«Ej. 

Moreover, if W = gjEj + g 2 E 2 + g 3 E 3 , we find, as one would expect, that 
V-W=f lgl +f 2g2 +f 3g3 . 

In future we shall not always trouble to point out when the pointwise 
principle is used to make definitions or deduce results about vector fields. So 
long as you are sufficiently familiar with the linear algebra of E 3 you should 
have no trouble with adapting this to vector fields. 



56 
Summary 

Notation 
V-W 

vx w 

livii 

E t , E 2 , E 3 
r, #, z 
p, #, <£ 



M334 II.6 



Page 82, line 1 

Text, page 55 

Page 82, line 2 

Page 82, Definition 6.1 

Page 82, Example 6.2.(1) 

Page 83, Example 6.2.(2) 



Definitions 

(i) Orthogonal vector fields 

(ii) Orthonormal vector fields 

(iii) Frame field E l5 E 2 , E 3 

(iv) Coordinate functions with respect to a frame field 



Text, page 55 
Text, page 55 
Page 82, Definition 6.1 
Page 83, Lemma 6.3 



Examples 

(i) The natural frame field U l5 U 2 , U 3 

(ii) The cylindrical frame field Ej, E 2 , E 3 
(iii) The spherical frame field F lf F 2 , F 3 



Page 82, second paragraph 
Page 82, Example 6.2.(1) 
Page 83, Example 6.2.(2) 



Page 82, Example 6.2.(1) 



Results 

(i) For the cylindrical frame field 

E x = cos# l^ + sintf U 2 

E 2 = - sin# l^ + costf U 2 

E 3 = U 3 . 
(ii) For the spherical frame field 

F t = cosip cos# L 1 ! + cos<p sin# U 2 + sirup U 3 

F 2 = -sin# Ui -i- cost? U 2 

F 3 = -sin</?cos# L 1 ! - sin<£sin# U 2 + cos</>U 3 . 

Page 83, Example 6.2.(2) 

(iii) If E l5 E 2 , E 3 is a frame field and V is a vector field, 
then 

V = (V-E^E! + (V-E 2 )E 2 + (V-E 3 )E 3 . Page 83, Lemma 6.3.(1) 

(iv) If E l5 E 2 , E 3 is a frame field, V = SfjEi and 
W = SgjE-, then 

V.W = Sf-g-. Page 83, Lemma 6.3.(2) 



Techniques 
(i) 



Page 82, Definition 6.1 



Recognition of frame fields, 
(ii) Production of frame fields using Gram-Schmidt and 

cross product (see Lemma ILL A). 

(iii) Expression of a vector field in terms of a frame 

field, and applications thereof. Page 82, Example 6.2 



M334 II.6 57 

Exercises 

Techniques (i) and (ii) 

1. Page 84, Exercise 1. 

Technique (Hi) 

2. Page 84, Exercise 2. 



Solutions 



w (p)* Lj.r v (p>- 



Page 84, Exercise 1. 

First we notice that ||V|| and j|W|| are never zero and so E 1 and E 2 are 
well-defined. Since V and W are linearly independent at each p, we know 
that 

V(p) * 

W(p) ¥= 

(W-V)(p ) 

IIV(p)|| 5 

V(p) =£ and so ||V|| is never zero. 
W = W- (W-E^Ej 

IIVH 2 
which is also never zero. 
We must show that E-E: = 5-. 



v_jv__ y^ _ iivii 2 
livifiivii" livil 2 "hvii 3 



E.-E, = — -• — = = — — = i • 



W W 

E2 * E2 = Mi' Mi ^ similarly ' 

E . E - = _X_ . _W_ 3 V-W = V-(W- W-EtEt) 
r 2 " IIVH ||W|| = iivii ||W|| IIVH iiwii 

_ V-W- (W-E^fV-E,) _ ||V||(E r W) - (W-EOfllVHErEQ 
IIVH ||W|| hvii iiwii 

IIVH (W-EQ - [|V||(W-E 1 ) _ 
IIVH iiwii 

Since E 3 = E 1 X E 2 , £^£3 = E 2 «E 3 = and 

E 3 -E 3 =((E 1 -E 1 )(E 2 .E 2 )-(E 1 .E 2 ) 2 ) 2 , by Lemma II. 1.8, 

= 1. 

Page 84, Exercise 2. 

(a) Ux = cos# E t - sin# E 2 

= cosip cos?? Fj - sin# F 2 - sirup cost? F 3 . 

(b) cost? Uj + sin# U 2 + U 3 = Ej + E 3 

= (cos<p + sin^F^ + (cos<£ - sin</?)F 3 



58 M334 II.6 

(c) (i) xUi + yU 2 + zU 3 = r cost? Ux + r sint? U 2 + zU 3 

= rE, + z U 3 ; 

(ii) xUx + yU 2 + zU 3 = p cos<p cost? Ui+ p cosy sint? U 2 + p simp U 3 

= pFi. 

If you cannot do this type of exercise by inspection, you may find the 
following method helpful. We can write the equations giving the Fj in terms 
of the Uj in matrix form as 



Fj \ /cos <p cost? cos <p sin t? simp \ lUi 



F 2 I = 
F 3 



- sint? cost? . 

- sirup cost? - simp sint? cos<p 



U 2 



= M 



u 2 
u 3 



Since the Fj and the Uj are both frames, M must be an orthogonal matrix, 



i.e. M 



1 = t M, so we can write /U/ 

U 2 
\U 3 , 



= t M IF A , giving the Uj in terms of the F-. 
F 2 



As an example, we work part (b)(ii): 
cost? Ui + sint? U 2 + U 3 = (cost?, sint?, 1) 



Ui\ 

u 2 
u 3 



= (cost?, sint?, l)tM /Fi 
F 2 

h 

= (cost?, sint?, 1) / cos<p cost? -sin t? - simp cost? 
cos<p sin t? cos i? - sirup sin t? 
sirup cos<p 

= (cosip + simp, 0, cos<p - sin<p) I F x \ 

F 2 



\ 



Fi 
F 2 
F 3 



= (cosip + simp)Fx + (cos<p - sin<p)F 3 



M334 II.7 59 

H.7 CONNECTION FORMS 

Introduction 

These last two sections of Chapter II are of quite a different nature to the preceding 
ones. The material is more abstract, and concerned almost entirely with forms. We 
expect you will find these sections difficult. However, they are not needed again 
until Chapter VI, so you may be content with just a superficial understanding at this 
stage of the course. If necessary, you can revise these sections when you reach 
Chapter VI: you should find them easier going at that reading after the lapse of 
time. 

Section II. 7 draws together the previous two sections and Section 1.5. Its purpose is 
to generalize the Frenet formulas to arbitrary frame fields. The matrix techniques 
introduced look messy but are simple to use in practice. 

READ: Section II. 7 (pages 85-90). 

Comment 

(i) Page 88, Theorem 7.3 The ij-th entry of dA*A = 2(dA)., (*A) . = 

k *k kj 

2 (dajj c )(a:j c ), which we write as Sa^da^ to keep to the standard con- 

IV IV 

vention of putting the differentials on the right. We cannot use such a 
conventional arrangement with the matrices dA and t A, however, as matrix 
multiplication is not commutative. 

Supplementary Comment 

(i) Page 89: dA The fact that d(cos #) = -sin § d#, and the other similar 

results, follow from Lemma 1.5.7. 

Summary 

Notation 

"ij Page 85, Lemma 7.1 

w Page 87, line 4 

A Page 88, line 11 

dA Page 88, line 18 

Definitions 

(i) Connection forms c^j Page 85, Lemma 7.1 

(ii) Attitude matrix (A) of a frame field Page 88, line 11 
(iii) Differential of a matrix whose entries are functions 

from E 3 to R Page 88, line 18 

Results 

(i) cojj is a 1-form and cjjj =-wji. Page 85, Lemma 7.1 

(ii) The connection equations: V^ = Scojj (V)Ej. Page 86, Theorem 7.2 

(iii) 03 = dA*A. p age 88, Theorem 7.3 

Technique 

(i) Calculation of the connection forms using 03 = dA^. 

Page 88, Theorem 7.3 



60 
Exercises 



M334 II.7 



Technique (i) 

1. Page 91, Exercise 4, using A and co. 

Theory Exercise 

2. Page 91, Exercise 5. 



Solutions 

1. Page 91, Exercise 4. 

A= j cos <p cost? cos <p sin # sin<p 

-sin i? cos# 

,-sintp cost? -simp sin# cos<p 



dA = 



cos <p d<p \ 


sin<pd<p / 



- sin <p cos # d<p - cos <p sin # d# - sin<p sin i? dtp + cos <p cos # d# 

-cos#d# -sin#d# 

' - cos <p cos # dip + sin <p sin # d# - cos <p sin # dy? - sin <p cos # d# 
co = dAtA = | cos<p d& d<p \ 

-cos<pd# sin<pd# 

-dip -simp d# 

so oj 12 = cos<p d# 
co 13 = d<p 
co 2 3 = sin<p d#. 

2. Page 91, Exercise 5. 

i 

= 2 V v (f i E i ), by Corollary 11.5.4.(1), 

i 

= 2 (Vtfj] Ej + fjVyEi), by Corollary 11.5.4.(3), 

i 

= 2 (V[fj] Ej + fj? co--(V)E-), by the connection equations, 
i J 

= 2 V[f:]E:+S 2 f i w i :(V)E:, by rearrangement, 
i J J j i J J 

= 2 {V[fj] + 2 f i co ij (V)}E j . 

In the case of the natural frame field all the connection forms are zero (prove!) 
and so this equation reduces to 

V V W = 2V[fj]Uj 

as given on page 79. 



M334 II.8 61 

II.8 THE STRUCTURAL EQUATIONS 

Introduction 

As mentioned in the previous section, you will probably find this section rather 
abstract and hard going. Do not attempt it until you feel fairly confident about both 
Section II.6 and Section II. 7. You should also be familiar with the ideas of linear 
functional, dual space and dual basis (see Unit M201 12, Linear Functional and 
Duality, Sections 1 and 2). 

This section deals with finding the exterior derivatives of the connection forms 
introduced in Section II. 7 and of the dual 1-forms of a frame field. As we do not yet 
know what an exterior derivative is, this section, like 1.6, must be approached purely 
formally, as manipulation of strings of symbols. In Chapter IV we shall find a 
meaning for the exterior derivative, and in Chapter VI we shall relate that to the 
results developed here. 

The results of this section are not needed until Chapter VI. You will probably find it 
hard: if so, do not worry about understanding it in depth now, but return to it from 
time to time so that it becomes fairly familiar before Chapter VI is reached. 

READ: Section II.8 (pages 91-95). 

Comments 

(i) Page 91: dual 1-forms In Lemma II.6.3 we saw that if Ej , E 2 , E 3 is a 

frame field then any vector field V can be expressed as 

V = (V-EJE! + (V.E 2 )E 2 + (V-E 3 )E 3 . 

In a similar manner, we obtain the fact that, if v is a tangent vector, 

V P = ( v p -Ei(p))Ei(p) + (v p -E 2 ( P ))E 2 (p) + (v p .E 3 (p))E 3 (p), 

simply by applying Lemma II.1.5 to the frame E x (p), E 2 (p), E 3 (p). Thus if 
we define the functions 0j from the set of all tangent vectors to E 3 to R (and 
extend to vector fields by the pointwise principle) by 

i(v p ) = Vp.Ei(p) 

we may write 

v p = MvpJE^p) + 2 (v p )E 2 ( P ) + 3 (v p )E 3 (p) 

V = 0,(V)E, + 2 (V)E 2 + 3 (V)E 3 . 

0i is certainly linear at each point, and so 0j is a 1-form. In fact, since the Ej 
are orthonormal, we have 

«i(Ej) = «ij 

which is precisely what is meant by saying that the 0j are the dual basis to 
the Ej. This is why we call the 0; the dual 1-forms of the frame field. 

(ii) Page 94: Example The 0; have not been found yet, but a method for so 
doing is given on page 95. The cojj were computed in Exercise II.7.4. 



62 
Summary 

Notation 
d: 



M334 II.8 



Page 91, Definition 8.1 



Definition 

(i) Dual 1-form { . 



Page 91, Definition 8.1 



Results 

(i) 

(ii) 
(iii) 

(iv) 
(v) 



If V is a vector field, then V = S0 i (V)E i . 

*i(Ej) = 5 ij. 

If is a 1-form, then = S0(E i )0j. 

If Ej = S aj-U., then 0j = 2 ajj dxj. 

J J 

The structural equations 






Techniques 

(i) Finding the dual 1 -forms 0} by adhoc methods 

or using the attitude matrix. 

(ii) Using the structural equations. 



Text, page 6 1 
Text, page 6 1 
Page 92, Lemma 8.2 
Page 92, line 20 



Page 92, Theorem 8.3 



Page 92, line 20 
Page 92, Theorem 8.3 



Exercises 

Technique (i) 

1. Page 96, Exercise 4. Part (c) depends on knowing what 9f/dr, 9f/8# mean 

and on knowing the formula 
3f 3f n 3f 

df = -dr + — d# +— dz. 

9r o# oz 

These will be explained in the final section of this text, so you may like to 
defer this part of the exercise until then. 



Technique (ii) 
2. 



Check the structural equations for the cylindrical frame field. (The con- 
nection forms were calculated on pages 89-90.) 



Theory Exercise 

3. Page 95, Exercise 1. 



M334 II.8 
Solutions 

1. Page 96, Exercise 4. 

(a) E x = cost? L^ +sin# U 2 

E. 2 =-sint? JJi + cost? U 2 
E 3 = U 3 . 



63 



By definition, ^(U:) = U-Ei = E^IL, so 

1 (U 1 ) = cost> 1 (U 2 )=sint? 0i(U 3 ) = O. 

We compare this with the value of dr on each of the U-. 

dr_ 

>X: 



dr(U i ) = U i [r] = 

1 l o: 



by Lemma 1.3.2. 




2\2 



SO 



Thus 



r = (x 2 + y 2 ) 



dx ( x 2 +y 2 )2 r 



= — = cost? 



3r 



9y (x 2 +y 2 ) 2 r 



L=-=sin# 



dr 

— = 0. 

3z 



dr(U 1 ) = cosi? = 1 (U 1 ) 
dr(U 2 ) = sin# = 2 (U 2 ) 
dr(U 3 ) = = 3 (U 3 ). 

Since the 1 -forms ^ and dr are equal on the frame field Uj they are 
equal on all tangent vectors and vector fields (by Lemma 1.5.4), and 
so 0j = dr. 

Similarly 2 (U-) = IL-E 2 , so 

02(Ui) = -sin i? 2 (U 2 ) = cos t$ d 2 (U 3 ) = 0. 



Also 



dd(U i ) = U i [d] =— . 

OX; 



i> = tan" 1 ] — 
x 



64 



so 











M334 II.8 


a#_ l ,-y 


-y 


-r sin# 
r 2 


-sini? 
r 




3x |y\ 2 x 2 


x 2 + y 2 




3.5_ 1 Jl\ 


x 


r cos # 

r 2 


cos # 

r 




ay 1+ |y|*W 


X 2 + y 2 




3z 











Thus . a 

r sin tf 

rdt?(U 1 ) = —=62(1],) 

r cos 1? ^ , 
r d#(U 2 ) = = 2 (U 2 ) 

r d#(U 3 ) = = 2 (U 3 ) 

and so 2 = r d#. 

3 (Ui) = 3 (U 2 ) = 3 (U 3 ) = 1 

dz(Ui) = dz(U 2 ) = dz(U 3 ) = 1 

and so 3 = dz. 

(b) Ejr] =dr(E i ) = 1 (E i ) = 5 li . 

2 (Ei) 5 2 i 
Ei [tf] =dt?(Ei)=-^ = T . 

E^z] = dz(Ei) = 3 (Ei) = 5 3i . 

(c) Ejf] = df(Ei) = I — dr + — d# + — dz (Ej) by the theorem 

to be proved in the final section, 

3f . 1 df . 3f 
= 5„ + - _ 6 2 i + _ 5 3 i. 
3r r ov dz 

Notice how we have used the Kronecker delta in (b) and (c) to 
convey several pieces of information in one statement. It does not 
matter if your solution did not use it, but bear in mind that you can 
often shorten the amount of writing by using it. 

Now we check the structural equations. 

X = dr, 2 = rd#, 3 = dz, 

co 12 = d#, co 13 = 0, co 23 =0 as found in Section II.7: 

d0! = d(dr) = O and w 12 a 2 + co 13 a 3 = d& a rd# + = 0; 

d0 2 = d(rd#) = dr a d# and co 21 a B x + co 23 a 3 = -dd a dr + = dr a d#; 

d0 3 = d(dz) = and co 31 aM co 32 a 2 = 0; 

dco 12 = d(d#) = and co 13 a co 32 = 0; 

dco 13 = and C0i 2 a co 23 = 0; 

dco 23 = and co 21 a co 13 = 0. 



M334 II.8 

3. Page 95, Exercise 1. 

d0 = d(2 fj0i) 



65 



= 2 d(f^i) 

i 

= 2 (dfi a 0. + fjd^j), 
i 

= 2 (d^ A fl. + f; 2 CO" a 0.), 
i J 

= 2 dfj a 0j + 2 2 qcoy- a 0j, 

= 2 {df: + 2 !>••} A 0, 

j i J J 



by Theorem 1.6.4.(2), 



by Theorem 11.8.3.(1), 



by rearrangement, 



66 



M334 II 



PARTIAL DIFFERENTIATION 

We have met two new coordinate systems in this chapter: r, d, z and p, d, <£. If you 
have met these before you may have come across expressions such as 9f/9r, 9f/9# 
and equations such as 

9f 9f 9f 

df = — dp + — d# + — dip. 

dp 9# 9<p 

Since the only partial derivatives we have so far formally defined are the 9f/3x-, 
where x- are the natural coordinate functions, our purpose here is to give a meaning 
to expressions such as 9f/9r, 9f/9# which is consistent with our idea of a partial 
derivative. 

To do this properly, we must first of all be quite clear about the use of the symbols 
x, y, z, as defined on page 4. They denote not "dummy variables", but the natural 
coordinate function from E 3 to R. That is, if p = (p 2 , p 2 , P3 ) then x(p) = pi , y(p) = 
P2> z (p) = P3- In particular (x, y, z)(p) = (pj , p 2 , p 3 ) = p, and so (x, y, z) is the 
identity function on E 3 . 

Now suppose that f is a function from E 3 to R. We must first notice that, although f 
is a single function from E 3 to R, it has different expressions in terms of x, y, z and 
in terms of p, #, <p say. For example, let us consider the function f = x 2 + y 2 + z 2 . We 
know p 2 = x 2 + y 2 + z 2 , so we can also write f = p 2 . Now, because (x, y, z) is the 
identity function, we can write 

f = f o ( x , y, z) = f(x, y, z) 

but f is not equal to f(p, #, <p) because 

f(p, 0, <p) = f o (p, 0, J) = p 2 + # 2 + <p 2 . 

Can we find a function g such that f = g(p, #, <p), that is f = gom where m is the 



function (p, #, <p): E' 



E 3 ? 



E ; 



m= (p, #,</>) 




In this case we can do so quite simply. If p € E 3 then 
f(p) = P 2 (P) = (P(P)) 2 

g (p, T>, ^)(p) = g(p( P ), 0(p), spi$)) 

and so we require the function g to pick out the first coordinate and square it: but 
the function that does this is precisely the function we have called x 2 . Thus g = x 2 
and so 

f =x 2 ° m. 
We return to the general case, and recall that the definition of 9f/9x is 



9f d 

-T-(P) = -j- f (t> P* Ps) 
ox dt 



, = ^7 f(t, y(p), 4p)) 

t=pj dt 



t=x(p). 



M334 II 67 

Roughly speaking, we are holding y and z fixed and differentiating with respect to x. 
If we want to define 3f/3p in a similar fashion we need to hold # and <p fixed and 
differentiate with respect to p. This is where we need to use the function g: since 

f = g(p, &, if) 

3f 
we define — by 
dp 



3f d 

^(p) = dT g(t, ^p),^(p)) 



t=p(p)- 



But this is precisely 3g/3x evaluated at (p(p), #(p), <^(p)), because, roughly speaking, 
3g/3x means "differentiate with respect to the first variable" no matter what the 
"variables" may be called. 

This is formalized in the following definition. 



Definition 

If m = (m lf m 2 , m 3 ) is a differentiable one-to-one function from a subset of E 3 to 
E 3 with differentiable inverse m" 1 , and f is a differentiable function from E 3 to R, 
and g is a function such that f = g o m (in fact g must be f ° m" 1 ), then 

3f 3g 

— (p) = £ (m(p)). 

dm- ox- 

Notice that this, definition of 3f/3mj is dependent on the whole function m: if m = 
(in! , m 2 , m 3 ) then 8f/3m^ may be different depending on whether m or m is being 
used. 

In practice this definition is very easy to work with. In our previous example we had 
f = P 2 , g = x 2 , and so 3f/3p = (3g/3x)(m)= 2x(p, &, \p) = 2p. Another example: if 
f = p cos if then g = x cos z, and 

3f = 3g 

(m) = cos z(p, &, <p) = cos ip; 
op ox 

3f 3 g/ 

— =-(m =0; 
3t? 3y V } 

3f 3g. . 

— -■— (m) =-xsinz (p, r?, <^)=- psimp. 
o<p oz 

This should show you that this process effectively treats p, #, <p as "dummy variables" 
just as x, y, z are normally treated, and so the new partial derivatives can be written 
down without the step of introducing g. 

For the usual partial derivatives 3f/3x, 3f/3y, 3f/3z we had the result (Corollary 
1.5.5): 

df =— dx + — dy + — dz. 
ox 3y 3z 

Now that we have defined other partial derivatives suitably, we can prove an 
analogous theorem. 



68 M334 II 

Theorem 

If m = (mi , m 2 , m 3 ) is a differentiable one-to-one function from a subset of E 3 to 
E 3 with differentiable inverse m" 1 , and f is a differentiable function from E 3 to R, 

then 

3f 
df = 2— dm-. 
bm i 1 

In particular, 

df dr+ - d# + -dz; 

3r ou oz 

3f 3f in 3f J 

df = —dp + - d0 + — d*. 
3p 3# 3<p 

(Corollary 1.5.5 is the particular case of this theorem for mj = Xj.) 



Proof 

We use the chain rule given in the additional text on Section 1.7. We can take 

g=f°m' 1 to obtain 

f = g o m. 
Now the chainrule gives 

If v is a tangent vector at the point p, this gives 



U v ) = g *( m (p)) m *( v )> 



(*) 



where g is written as a matrix evaluated at m(p) and the vectors f^v), m^v) are 
written as column vectors. Now, Theorem 1.7.5 tells us that 

f # (v) = (v[f]) = (df(v)): 

this is a 1 X 1 matrix so that corresponding column vector is also (df(v)). The same 
theorem tells us that 

m *( v ) = ( v t m i] > v t m 2l » v t m 3] ) 

= (dm^v), dm 2 (v), dm 3 (v)), 

which is 

'dmi(v)| 
dm 2 (v) 



as a column vector. 



By Corollary 1.7.7 g^has matrix 

'9g 9g_ dg \ 
k 9xi' 3x 2 ' 3x 3 | 



M334 II 

Evaluating this at the point m(p) we have 



\ 

Ji (m(p)), b ± (m(p)), *J- (m(p)) 
OX! dx 2 ox 3 



69 



which is precisely 



3f 3f 3f 

om l 3m 2 am 3 ' 

by the definition of (df/8m-). Putting the expressions we have obtained into (*): 



(df(v)) = 



af af af 

x~ (p)> r— (p), r — (p) 
onij dm 2 om 3 



dm^v) 
dm 2 (v) 
dm 3 (v) 



9f 3f 3f 

— (p)dm,(v) + — (p)dm 2 (v) + _ (p)dm 3 (v) 

om! 3m 2 3m 3 v ' 

3f 9f 3f \ \ 

= -— dmj + -— dm 2 + - — dm 3 (v) 
\om l dm 2 om 3 | 

This is true for all tangent vectors v, and so 

3f 3f 3f 

df = dm» + — dm 2 + dm 3 . 

m! om 2 om 3 

You do not need to know any details of this proof. All you need for this course is 
the defintion of general partial derivatives and the result stated in the theorem. 



Summary 

Notation 

[elf 3f 3fl 
l3r' 3#' 3zJ 

(df df 3fl 
|3p'3#' dd 



Text, page 66 
Text, page 66 



Definition 

(i) Partial derivatives 



Text, page 67 



Results 

af 3f 3f 

(I) df=~dr + — d#+ — dz. 

3r 3# 3z 

af 3f 3f 

(II) df = --dp+— d^+-~d«^. 

op 3# dip 



Text, page 68 
Text, page 68 



70 

Techniques 

(i) Finding partial derivatives. 

(ii) Using the expressions for df in terms of the 
partial derivatives. 



M334 II 



Text, page 67 
Text, page 68 



Exercises 



Technique (i) 

3f 3f df 

1. Find — , — , — -when f = p costf cos </?. 

dp 3# by 

df 3f 3f , r . , 

2. Find — ; — , — when f = x -4 + ir. 

3r 30 3z 



Technique (ii) 

This has already been tested in Exercise 11.8.4(c). Do that exercise now if you 
previously deferred it. 



Solutions 

1. f = pcos# COS<£ 

= x cos y cos z (p, #, if) 

= g(p, #, <p) where g = x cos y cos z. 

— = — (p, #, \p) = cosy cosz (p, #, <p) = cost? cos<p. 
3p 3x 

— = — (p, #, tp) = -x siny cos z (p, d, <p) = - p sin# cos (p. 
3# 3y 

— = — (p, #, <p) =-x cosy sinz (p, #, </?) = - p cos# sin<p. 
3<p 3z 

2. We must first of all express f in terms of r, &, z. 




From the diagram we see that 
x = r cos & 



M334II 71 

and so 

f=x 2 + z 2 

= r 2 cos 2 & + z 2 . 

As before, we can introduce g = x 2 cos 2 y + z 2 , or we can go directly to the 
results: 

9f 

— = 2r cos 2 i?, 
3r 

3f 

— = -2r*cosi? sin#; 

3f 

A" =2z - 
oz 



72 M334 II 

FURTHER EXERCISES AND SOLUTIONS 

Section II. 1 

Recommended Exercises 

Page 49, Exercise 2. This shows that Euclidean distance is a metric in the sense of 
M202andM231. 

Page 50, Exercise 11. This relates the vector operation grad to the norm and dot 
product. 

Solutions 

2(a). d(p, q) = ||p - q|| = (2(p- - qj) 2 ) 2 , which is non-negative. 

d(p, q) = => 2(pj - qj) 2 = 0, and a sum of squares is zero if and only if each term 
is zero, that is pj = qj. 

2(b). ( Pi - qj) 2 = (qj - Pi ) 2 so d(p, q) = d(q, p). 

2(c). d(p, q) + d(q, r) = ||p - q|| + ||q - r|| > ||p - q + q - r|| = j|p - r|| = d(p, r). 

11(a). df(v) = 2 Vi _l = (Sarins -i Ui(p)) = vVf(p). 
3xy 3xi 

11(b). Putw= (Vf)(p). 

u[f] = u-Vf(p) = u«w 

and lu-wl < INI ||w|| = ||w|| 

with equality when u = -5^- . 

w 



Section II. 2 

Technique Exercise 
Page 55, Exercise 1. 

Other Recommended Exercises 

Page 55, Exercise 5. This deal with unit-speed repararmetrizations in the sense of 

Theorem II. 2.1. 

Page 56, Exercise 10. This shows that arc-length is preserved by reparametrizations 
which are symmetric in the sense of Comment (i). 

Solutions 

1(a). (2,2^, (2, 2, 1) (2) 1; ,- 

t 2 + 2,3; 

<°' 2 > 2t W (0 ' 2 ' 2) (2,U)- 



M334 II 73 

1(b). s(t) = 2t + |;p 

5. If /3j is the unit-speed reparametrization based at t = tj, and sj is the arc- 

length function based at t = tj, 

«(t)=^i(s 1 (t))=^ 2 (s 2 (t)). 
Si(t) = j v(u)du 

ft 2 ft 

= J v(u)du + I v(u)du 

= s + s 2 (t) 
where s is the arc-length of a from t t to t 2 . 

10. Put0 = a(h). 

(a) Arc-length of from a to b 

b 



J v 0( u ) du 



V(d) 



= J h'(u)v a (h(u))du 

h-'(c) 

= J v a (t)dt 
c 

= arc-length of a from c to d. 



Arc-length of |3 from a to b 


= J V£(u)du 
a 


fh _1 (c) 
= J (-h'(u))v a (h(u))du 
h" 1 (d) 


= -J v a (t)dt 
d 


= f v^tjdt. 
c 



74 M334 II 

Section II.3 

Technique Exercises 

Page 63-64, Exercises 2, 3. 

Other Recommended Exercises 

Page 64, Exercise 5. This expresses all the Frerret formulas in a single format. 

Page 64, Exercise 6. This deals with the approximation of a curve by part of a circle. 

Solutions 

(l+s)*-(l-s)* 1 



2. T(s) = 

N(s) = 
B(s) =1 



2 V2 



{ 22T (H> a ,o 



\ V* ' V2 f 

-(1+S)2 (1-S) 5 1 

T ' T 'V2| 



k(s) = r(s) = _^___ i 
W V2(l-s 2 ) 2 

3. Evaluate both sides of each equation. 

5. AXT = kBXT = kN 

A X N = tT X N + kB X N = tB - kT 
AX B = rTX B=-rN. 

6. The only solution for 7 is 

s s 

7(s) = c - r cos - N + r sin —T 
r r 

Nn 
where c = — + j3(0) 
/c 

1 
r = 

Section II.4 

Technique Exercises 

Pages 74-75, Exercises 1(a), 3(a), 4 (first part), 6, 7, 9(a), (b). 

Other Recommended Exercises 

Page 74, Exercise 5. This gives a formula for K. 

Page 75, Exercise 13. This introduces another special curve derived from a given 
curve. Since one of a, a is negative, you will first have to adapt Example II.3.3 for 
the case where a is negative. 



M334 II 75 

Solutions 



1 

2+t 2 
1 



!( a )' T(t)=— 2 (2,2t,t a ) 



N(t) = — 2 (-2t,2-t 2 ,2t) 

B(t)=-^L(t 2 ,-2t,2) 
2 



fc(t) = r(t) = 
3(a). T(0) = 



(2+t 2 ) 2 * 
(1,0,1) 



V2 
N(0) = (0, 1, 0) 

B(0) = tl^il 

K(0) = 1 

3 

T(0)=-. 

4. Evaluate both sides of each equation. 

6. a' = cT. 

a" = cT' = c 2 /cN. 
a'x a" = c 3 /cB. 
Now substitute in Theorem II.4.3. 

7. T(t)=|-^sint,^cost,|) 

N(t) = (-cost, -sint, 0) 

, lb b a\ 

B(t) = - sin t, — cos t, — 

\c c c/ 



«(t) =i 2 



r(t)=K 



9( a). B JL2d> 
V2 

#=- 

4 



7(t) = ( t -| tVt+ |). 



9(b). u = (0,0,1) 

* = * 
4 

7(t) = (3t - t 3 , 3t 2 , 0). 



76 M334II 

dv _ 

5. a" = — T + kv 2 N 

dt 

ll«"|| 2 =[~| 2 +(/CV 2 ) 2 . 

3. S S 

13. Ifa>0, then fc(s) = -=, N(s) = (- cos-,- sin-, 0); 

c c c 

if a < 0, then k(s) = — «, N(s) = (cos-, sin-, 0). 
c c c 

Substitute these to obtain 

0ab* = "ah 
for all a. Now (a) = a so 

<%b* ='<W 



Section II. 5 

Technique Exercises 

Page 80, Exercises 1(b), 2(b), (d), (e), (f), 4. 

Other Recommended Exercises 

Page 80, Exercise 3, This extends the result that if Y is a vector field on a curve and 

11Y|| is constant then Y-Y' = 0. 

Page 80, Exercise 6(b). This deals with the definition of covariant derivative. 

Page 80, Exercise 7. This exercise introduces the bracket of two vector fields. 

Solutions 

1(b). (l,2,4) p 

2(b). -yU 3 . 

2(d). -sinx U! + cosx U 3 . 

2(e). -y 2 cosx \J X - y 2 sin x U 2 . 

2(f). (y 2 - xcosx -yzsinx)U! + (-xsinx +yzcosx)U 2 - 2xy U 3 . 

4. X = Sxj Uj. 

V v X = 2 V [ Xi ] Uj = S 2 v: -^ Ui = S 2 Vj 5 ij Ui 

v . i i . j a Xj i j J J 



-ZviU-V. 



W(p+tv)-W(p+tv) = constant 
=»W(p+tv)-W(p+tv)' = 
=>W(p)-V v W = 0. 



7(a). Both sides equal X 2 

i J 



dw: dv- 

v j — " w j — 
3 Xj dx- 



M334 II 77 

6(b). W(a)'(t) = V a , (t) W = V v W. 

* 3f 

Bx: 

J / 1 

7(b). The definition is antisymmetric. 

7(c). [U,[V,W] ] [f] = U[V,W] [f] - [V,W]U[f] 

= UVW[f] - UWV[f] - VWU[f] +WVU[f]. 
Adding the other two similar expressions to this gives zero. 

7(d). [fV,gW] [h] = fV[gW[h] ] - g W[fV[h] ] 

= f(V[g]W[h] +gVW[h])- g(W[f]V[h] + fWV[h]) 
= fV[g] W[h] - gW[f] V[h] + fg[V,W] [h]. 

Section II. 6 

Techinque Exercises 

Pages 84-85, Exercises 3, 4. 

Solutions 

3. Choose any unit E 2 orthogonal to E 1 and put E 3 = Ej X E 2 . For example 

E 2 = sin z U 2 - cos z U 3 

E 3 = - sin x Uj + cos x cos z U 2 + cos x sin z U 3 . 

4. The same formulas are obtained as for the spherical frame field. 

Section II. 7 

Technique Exercises 

Pages 90-91, Exercises 1, 2, 3, 7. 

Solutions 

df 

1. <^i2-0;o; 13 = co 23 = - 7 -. 

2. All zero. 

3. A is orthogonal. 
oj 12 = -df; 
co 13 = cosf df; 
co 23 = sinf df. 



78 

7(a). For any f, 



M334 II 



Fi|fl =cosiz> cos v ^— + cos kd smj? — - + sin (/>»-. 
11 J ^ ox r 3v oz 



8y 



p = (x 2 + y 2 + z 2 ) 5 



t^tan- 1 y 
U/ 



<p = tan 



-l 



^(x 2 +y 2 ) 2 ' 
evaluate the partial derivatives and substitute. 

7(b). - sin p F 2 + cos p F 3 . 



Section II.8 

No further exercises are recommended. 



M334 II 79 

DIFFERENTIAL GEOMETRY 

I Calculus on Euclidean Space 

II Frame Fields 

III Euclidean Geometry 

IV Calculus on a Surface 

V Shape Operators 

VI Geometry of Surfaces in E 




4 PART III THE OPEN UNIVERSITY 



— 9 



is*- '<■->■-,*■:.;■■ ■ 

51" • Mathematics: A Third Level Course 



DIP 

PART III EUCLIDEAN GEOMETRY 



DIFFERENTIAL GEOMETRY 



9 



THE OPEN UNIVERSITY 
Mathematics: A Third Level Course 

DIFFERENTIAL GEOMETRY PART III 



EUCLIDEAN GEOMETRY 



A commentary on Chapter III of O'Neill's 

Elementary Differential Geometry 

Prepared by the Course Team 



THE OPEN UNIVERSITY PRESS 



Course Team 



M334 III 



Chairman: 



Mr. P.E.D. Strain 
Dr. R.A. Bailey 



Lecturer in Mathematics 
Course Assistant in Mathematics 



With assistance from: 



Dr. J.M. 
Mr. G.J. 
Dr. P.M. 
Mr. P.B. 
Dr. F.C. 
Mr. T.C. 
Mr. RJ. 
Dr. C.A. 
Mr. M.G. 



Aldous 
Burt 
Clark 
Cox 

Holroyd 
Lister 
Margolis 
Rowley 
, Simpson 



Senior Lecturer in Mathematics 
Lecturer in Educational Technology 
Lecturer in Physics 
Student Computing Service 
Lecturer in Mathematics 
Staff Tutor in Mathematics 
Staff Tutor in Mathematics 
Course Assistant in Mathematics 
Course Assistant in Mathematics 



Consultants: 



Prof. S. Robertson 
Prof. T. Willmore 



Professor of Pure Mathematics, 
University of Southampton 

Professor of Pure Mathematics, 
University of Durham 




The Open University Press, Walton Hall, Milton Keynes, MK7 6AA 

First publishedV^S^eprinted 1979 

Copyright © 1975 The Open University 

All rights reserved. No part of this work may be reproduced in any form, by mimeograph or any other means, 
without permission in writing from the publishers. 

Produced in Great Britain/by 
The Open University Pros 

ISBN 335 05702 0>y 

This text forms part of the correspondence element of an Open University Third Level Course. The complete list 
of parts in the course is given at the end of this text. 

For general availability of supporting material referred to in this text, please write to: Open University 
Educational Enterprises Limited, 12 Cofferidge Close, Stony Stratford, Milton Keynes, MK11 1BY, Great 
Britain. 

Further information on Open University courses, may be obtained from The Admissions Office, The Open 
University, P.O. Box 48, Milton Keynes, MK7 6AB. 



1.2 



M334 III 



CONTENTS Page 

Set Book 4 

Bibliography 4 

Conventions 4 

III. 1 Introduction and Isometries of E 3 5 

III. 2 The Derivative Map of an Isometry 14 

III. 3 Orientation 20 

III. 4 Euclidean Geometry 26 

III. 5 Congruence of Curves 32 

Further Exercises and Solutions 38 



4 M334 III 

Set Book 

Barrett O'Neill: Elementary Differential Geometry (Academic Press, 1966). It is 
essential to have this book: the course is based on it and will not make sense without 
it. 



Bibliography 

The set books for M201, M231 and MST 282 are referred to occasionally; they are 
useful but not essential. They are: 

D.L. Kreider, R.G. Kuller, D.R. Ostberg and F.W. Perkins: An Introduction to 
Linear Analysis (Addison- Wesley, 1966). 

E.D. Nering: Linear Algebra and Matrix Theory (John Wiley, 1970). 

M. Spivak: Calculus, paperback edition, (W.A. Banjamin/Addison-Wesley 1973). 

R.C. Smith and P. Smith: Mechanics, SI edition (John Wiley, 1972). 



Conventions 

Before starting work on this text, please read M334 Part Zero. Consult the Errata 
List and the Stop Press and make any necessary alterations for this chapter in the set 
book. 

Unreferenced pages and sections denote the set book. Otherwise 

O'Neill denotes the set book; 

Text denotes the correspondence text; 

KKOP denotes An Introduction to Linear Analysis by D.L. Kreider, R.G. 
Kuller, D.R. Ostberg and F.W. Perkins; 

Nering denotes Linear Algebra and Matrix Theory by E.D. Nering; 

Spivak denotes Calculus by M. Spivak; 

Smith denotes Mechanics by R.C. Smith and P. Smith. 

References to Open University Courses in Mathematics take the form: 
Unit Ml 00 22, Linear Algebra I 
Unit MST 281 10, Taylor Approximation 
Unit M201 16, Euclidean Spaces I: Inner Products 
Unit M231 2, Functions and Graphs 
Unit MST 282 1, Some Basic Tools. 



M334 III.l 5 

HI.1 INTRODUCTION AND ISOMETRIES OF E 3 

Introduction 

The aim of this chapter is to prove that a unit-speed curve in E 3 is determined, apart 
from its position in space, by its curvature and torsion functions, which were intro- 
duced in Section II. 3; in other words, that the curvature and torsion of a curve, as 
functions of the arc-length, tell us all there really is to know about it. We cannot 
begin to prove this until we have decided what we mean by "apart from position in 
space": when are two curves in different positions in E 3 going to be declared 
"essentially the same"? We shall not finally answer this question until Section III.5, 
but we make a start in this section by defining isometries, that is mappings of E 3 
which preserve distance. You have met some isometries before — translations in Unit 
M201 15, Affine Geometry and Convex Cones, Section 1.1, and orthogonal linear 
transformations in Unit M201 24, Orthogonal and Symmetric Transformations, 
Section 1. You should remind yourself of the results of the latter section now, as 
they are used in this section. Apart from that, this section depends only on Sections 
1.7 and II. 1; the later sections of Chapter II are not referred to yet. 

READ: Introduction to Chapter III and Section III.l (pages 98- 102). 

Comments 

(i) Isometries An important point about the isometries of E 3 is that they 

form a group; that is, that they satisfy the following conditions. 

(a) The composition of two isometries is again an isometry — this is 
Lemma 1.3. 

(b) Composition of isometries is associative — this is true for com- 
position of any functions. 

(c) There is an identity isometry I such that IF = FI = F for all iso- 
metries F — this is true because the identity mapping I is an 
isometry. 

(d) Each isometry F has an inverse F" 1 which is also an isometry. This 
condition requires slightly more checking. It is an immediate con- 
sequence of Definition 1.1 that isometries are one-to-one mappings, 
for if F(p) = F(q) then 

d(p, q) = d(F(p), F(q)) = d(F(p), F(p)) = 

and so p = q. The fact that isometries of E 3 are also onto mappings 
follows from Theorem 1.7, because the translation T is onto and the 
orthogonal transformation C is onto, so their composite F = TC must 
also be onto. Since F is one-to-one and onto F has an inverse func- 
tion F" 1 = C" 1 !" 1 , which is a mapping because both C" 1 and T" 1 are 
mappings (see Comments (ii) and (ill)). Now 

d ( F " 1 (p)»F- 1 (q)) = d(FF' 1 (p), FF _1 (q)) because F is an 

isometry 
= d(p, q) because FF" 1 = I 

and so F" 1 is indeed an isometry. 



6 M334 III.l 

(ii) Translations The translations form an important set of isometries, as 
noted in Example 1.2(1). We shall use the notation T a to denote translation 
by a, that is, 



Then 



T a(p) = P + a. 



Tt>T a (p) = p + a + b = T a + b (p) = T b + a (p) 



so T b T a is translation by b + a. In fact the translations correspond to the 
points of E 3 in such a way that the composition of translations corresponds 
to ordinary vector addition; hence the commutativity of composition of 
translations follows from the commutativity of vector addition. We have the 
following commutative diagram: 



a, b 



corresponding 
translation 



T a> T b 



■► a + b 



composition 



corresponding 
translation 



-► T a T b = T a + b 



Since the identity mapping I is translation through 0, T a = T_ a and we 
have this commutative diagram also: 



additive inverse 



corresponding 
translation 




corresponding 
translation 



(T a r = T_ a 



Thus the translations also form a group. 



(iii) Orthogonal Transformations As noted in Lemma 1.5, these form 
another important set of isometries. The definition given on page 100 is in 
fact equivalent to that given in Unit M201 24, Orthogonal and Symmetric 
Transformations, page 8, because all the spaces in which we are interested 
are finite-dimensional. Thus we can make use of the result proved in that 
unit, that the following conditions on a linear transformation C of finite- 
dimensional Euclidean spaces are all equivalent: 

(a) C preserves dot products (that is, C is an orthogonal transformation); 

(b) The rows of the matrix C are orthonormal; 



M334 III.1 7 

(c) The columns of the matrix C are orthonormal; 

(d) t CC = C*C = I; that is, C" 1 = *C (that is, C is an orthogonal matrix); 

(e) C maps some orthonormal basis to another orthonormal basis; 

(f) C maps all orthonormal bases to orthonormal bases. 
Since, for orthogonal transformations Q, Cj, C, 

(CiC 2 ) = C 2 C 1 = C 2 Ci = (CiCj) 

and 

\cr l ) = Yc) = C = (C 1 )' 1 

this immediately tells us that the composite of two orthogonal 
transformations is an orthogonal transformation and that the inverse of an 
orthogonal transformation is an orthogonal transformation. Thus these also 
form a group. 



(iv) Page 101: Theorem 1. 7 This breakdown of an isometry F as TC is very 
useful. In future work with isometries we shall nearly always use this ex- 
pression, sometimes without explicitly defining T and C. If F t = TiCi and 
F 2 = T 2 C 2 are isometries then their composite F t F 2 is also an isometry 
(Lemma 1.3) so we can write F 1 F 2 = T 3 C 3 , where T 3 is the translation part of 
FxF 2 and C 3 is the orthogonal part of FxF 2 . Then 

T3C3 = TiCiT 2 C 2 

but it is not generally true that T 3 = TjT 2 (although C 3 = CiC 2 ) because Ci 
and T 2 do not commute in general. (See the Theory Exercises if you are 
interested in investigating this further.) 



Supplementary Comments 

(i) Page 99: rotation The formulas for the coordinates of q in terms of 

those of p were worked out in Unit M201 24, Orthogonal and Symmetric 
Transformations, Section 1.2. 



(ii) Page 101: lines -10 and -9 If we put r = ap + bq, then rj = apj + bqj for 
i = 1, 2, 3. Using the identity established in line - 11, 

F(r) = 2riF(ui), F(p) = 2 Pi F(ui), F(q) = 2 qi F(ui) 



so 



F(r) = 2r i F(u i ) 

= 2 (a Pi + b qi )F(ui) 
= 2ap i F(u i ) + 2bq i F(u i ) 
= aZp i F(u i )+bSq i F(ui) 
= aF(p) + bF(q). 



8 
Summary 



M334 III.l 



Notation 

F 
T 
T 
C 
I 

Definitions 

(i) Isometry F 

(ii) Translation by a, T a 

(iii) Orthogonal transformation C 

(iv) Identity mapping I 

Example 

(i) Rotation about the z axis 

Results 

(i) 
(«) 



The composite of two isometries is again an 
isometry. 

For translations: 



(iii) 
(iv) 

(v) 

(vi) 

(vii) 



T a T b = T b T a = T a + b> 

(T a r 1= T- a . 

Given p, q in E 3 , there is a unique translation T 
such that T(p) = q. 

If T is a translation and T(p) = p for any p, then 
T = I. 

Translations are isometries. 

Orthogonal transformations are isometries. 

Any isometry fixing is an orthogonal trans- 
formation. 



(viii) Each isometry F has a unique expression as TC, 
where T is a translation and C is an orthogonal 
transformation. Moreover T = Tjr(o). 

(ix) Every isometry has an inverse isometry. 
Techniques 

(i) Recognition of isometries, translations, orthogonal 

transformations. 



(ii) Expression of an isometry in standard form TC. 

(iii) Calculation of the image of a point under an iso- 
metry expressed in standard form. 



Page 98, Definition 1.1 
Page 98, Example 1.2(1) 
Text, page 6 
Page 100, line 11 
Page 99, line -6 



Page 98, Definition 1.1 
Page 98, Example 1.2(1) 
Page 100, line 10 
Page 99, line -6 



Page 99, Example 1.2(2) 



Page 99, Lemma 1.3 

Page 99, Lemma 1.4(1) 
Page 100, Lemma 1.4(2) 

Page 101, Lemma 1.4(3) 

Page 100, lines 8-9 
Page 98, Example 1.2(1) 
Page 100, Lemma 1.5 

Page 100, Lemma 1.6 

Page 101, Theorem 1.7 
Text, page 5. 



Page 98, Definition 1.1, 
Page 98, Example 1.2(1), and 
Page 100, line 10. 

Page 101, Theorem 1.7 



Page 102, line -5 



M334 III.l 
Exercises 

Technique (i) 

1. Page 103, Exercise 4. 

2. Page 103, Exercise 6, (a) — (d), first part. 

Technique (ii) 

3. Page 103, Exercise 6, second part. 

Technique (Hi) 

4. Page 103, Exercise 5(a). 

Theory Exercises (omit if short of time) 

5. Page 103, Exercise 1. 

6. Page 103, Exercise 2. 

7. Page 103, Exercise 3. 

Solutions 



1. Page 103, Exercise 4. 



As noted in Comment (iii), C is orthogonal <=► C X C = I •*= > the rows of C are 
orthonormal. We check the last condition as it involves only six calculations. 

.11. lU-l 2 _J.Ll (4 + 4 + 1)=1 



2. 1 
3' 3' 



3' 3' 



| }, " 114(4 + 1 + 4) = ! 



\F' 3' 3J \3' 3' 3/ 9 (1 4 4) l 



2 2 
3' 3' 



2 1 
3' 3' 



|Ll ( - 4 + 2 + 2) = 



H-fl-Ml)=i (2 + 2 - 4) = () 



This shows that C is orthogonal. 



C(p) = 




v 



2 
3 

19 
3 

]_ 



10 



M334 III.l 



2. 



C(q) = 




lKR)(l)=-i(-10-76-49)=-iP = -15. 



p-q = 3.1 + 1.0 + (-6)f3) = 3 - 18 = - 15. 
C( P )-C(q) -(§)(- 4) + ji„ „., 3)l3) 

Page 103, Exercise 6. 

In each case we must check whether d(F(p), F(q)) = d(p, q) for all p and q. 

(a) F(p)=-p. 

d(F(p),F(q)) = d(-p,-q) 

= ||-p-(-q)|| 

= ||(-l)(p-q)|| 

= I" 1 I IIP" q|| (using ||ap|| = |a| ||p||) 

= i||p-q|| 

= d(p, q). 
This is true for all p and q, so F is an isometry. 
In this case F: pi yp-aa is just the perpendicular projection of E 3 



(b) 



onto the one-dimensional subspace spanned by a. We can easily see 
from the picture 




that F does not preserve norms: for example, we can choose p to be 
orthogonal to a and then F(p) = so ||F(p)|| ¥= ||p||. Since F does 
not preserve norms, F cannot preserve dot products. 

Alternatively we proceed as follows. 

d(F(p), F(q)) = d(p-a a, q-a a) 

= ||p-aa- q-aa|| 

= ||(p-a- q-a)a|| 

= |p*a- q*a| ||a||, as above, 

= |p*a- q*a|, since ||a|| = 1, 

= l(p-qH 

= l|p-q|INI|cost?|, 



where t? is the angle between 
p-q and a, 



= d(p, q) |cosi?| 



since Mai 



= 1. 



which is not equal to d(p, q) unless # is a multiple of it. 



M334III.1 11 

As yet another approach, since the image of F is not the whole of 
E 3 , F is not onto E 3 and so F cannot be an isometry, by Comment 

(i). 

(c) F(p) = (p 3 - l, P2 -2, Pl -3). 

d(F(p), F(q)) = d((p 3 - 1, p a " 2, Pl - 3), (q 3 - 1, q 2 " 2, qi - 3)) 
= ||(P3- l,p 2 -2,p 1 -3)-(q 3 - l,q 2 -2, qi -3)|| 

= ||(P3- q3»Pa- q 2> Pi- qi)|| 

= ((P3-q 3 ) 2 + (Pa-q 2 ) 2 + (pi-qi) 2 ) 1 

= ||(pi - qi» P2 - qa, p 3 - q3)|| 
= l|p-q|l 

= d(p, q). 
This is true for all'p and q, and so F is an isometry. 

(d) F(p) = ( Pl ,p 2 ,l). 

d(F(p), F(q)) = d(( Pl , p 2 , 1), (q lf q 2 , 1)) 

= ||(Pi> Pa> 1) " (qi, qa, 1)|| 
= ||(Pi- qi.Pa" q2, 0)|| 

= ((Pi-qi) 2 + (P2-qa) 2 ) 21 , 
which is not equal to d(p, q) unless p 3 = q 3 . For example, 

d((0, 0, 0), (0, 0, 1)) = 1 but 

d(F(0, 0, 0), F(0, 0, 1)) = d((0, 0, 1), (0, 0, 1)) = 0. 

Thus F is not an isometry. 

Again, we could note that, since the image of F is 
{ p G E 3 : p 3 = 1 } =£ E 3 , 

F is not onto, and so cannot be an isometry. 

3. Page 103, Exercise 6. 

(a) F(p) =_ p and so F = -I, which is an orthogonal transformation. 
This is already in standard form, and so T is the identity, C = -I. 

(b) F is not an isometry. 

(c) F(0) = (-1, -2, -3) so T, the translation part of F, is translation by 
(-1,-2,-3). If F = TC, 

F(p)=TC(p) 

i.e. (p 3 - l,p 2 - 2, Pl - 3) = C(p) + (-l,-2,-3) 

i.e. (p 3 , p 2 , Pl ) + (-1, -2,-3) = C(p) + (-1, -2, -3) 

so C(p) = (p 3 , p 2 , pj) 

/0 1^ 

so C has matrix 10 

10 

(In fact this represents reflection in the plane x = z.) 

(d) F is not an isometry. 



12 

4. 



M334 III.l 



5. 



Page 103, Exercise 5(a). 
F(p) = T a C(p)=a + C(p) 
/ 1 



V2 



-1 



V2 




/- 



V2 



10 

V2. 



1 - V2 



-1 + V2, 

Page 103, Exercise 1. 

Put F = CT a . Since C and T a are isometries F is an isometry (by Lemma 
III.l. 3), and thus by Theorem III.l. 7 we can write F = T^D where T. is 
translation through b and D is an orthogonal transformation. Now 

CT a (p) = T b D(p) for all p 



i.e. 



i.e. 



C(a + p) = b + D(p) 



C(a) + C(p) = b + D(p). 
Putting p = gives C(a) = b because C(0) = D(0) = 0. 
Substituting back, we have 



i.e. 



and so 



Thus 



b + C(p) = b + D(p) 

C(p) = D(p) 

C = D. 

F = T b D = T C ( a )C. 



for all p 
for all p 



M334 III.1 



13 



Alternatively, we could show that CT a and T£/ a \C have the same effect on 
every point of E 3 . For p in E 3 , 

CT a (p) = C(a + p) = C(a) + C(p), 

while 

T C ( a )C(p) = T C(a) (C(p)) = C(a) + C(p). 

6. Page 103, Exercise 2. 

F = T a A and G = T b B, so 

FG = T a AT b B. 
By the result of Exercise III. 1.1, 

AT b = T A(b) A 
and so 

FG = T a AT b B = T a T A(b) AB. 

T a T A(b) is the translation through a + A(b); AB is an orthogonal trans- 
formation, being the product of two orthogonal transformations. Thus FG 
has translation part T a + 'A(b) anc ^ orthogonal part AB. 

Similarly GF has translation part T b + jj/ a \ and orthogonal part BA. 

7. Page 103, Exercise 3. 

F = T a C. Since T a and C both have inverses we know that 

F" 1 = (TaC)' 1 = C' 1 ^ 1 . 
Now 

C^V 1 = C _1 T_ a , by Lemma 111.1.4(2), 

= Tc-i(_ a )C _1 , by Exercise III. 1.1, 
= T-^-i/^C" 1 , by linearity of C" 1 . 
Thus the translation part of F" 1 is T_c;-i( a ); the orthogonal part of F _1 is C" 1 . 



14 



M334 III.2 



III.2 THE DERIVATIVE MAP OF AN ISOMETRY 



Introduction 

This section follows on from Sections III.l and 1.7. We also need to use certain 
results from M201 — that there exists a unique linear transformation taking one 
basis of E 3 to another basis of E 3 (see Unit M201 2, Linear Transformations, 
Section 1.8) and that a linear transformation is orthogonal if and only if it takes an 
orthonormal basis to another orthonormal basis (see Unit M201 24, Orthogonal and 
Symmetric Transformations, Section 1.2). 

The purpose of this short section is to show that there is a unique isometry taking 
any given frame in E 3 to any other given frame in E 3 . In order to do this we have to 
look at the derivative map F*: we find that this takes a particularly simple form 
when F is an isometry, for if F = TC then F* is represented simply by C. 

READ: Section III. 2 (pages 104- 106). 



Comment 



(i) Page 105: lines -18 to -15 



F^p and C cannot be the same function, for 



they act on different spaces — T p (E?Ji andE 3 respectively. However, T p (E 3 ) is 

isomorphic to E 3 via the "canonical" isomorphism D : E 3 ► T D (E3) given 

by P P 



-»V T 



(v G E 3 ). 



3» 



To say that F^ and C "differ only by the canonical isomorphisms of E 
means, in effect, that the following diagram commutes; that is, that F +p and 
C send vectors that correspond to each other via 0« to vectors that corres- 
pond to each other via <b„, x . 

F(P) 




* T F(P) (E3 > 




► F *p( v p) = Cv 



F(P) 



Supplementary Comment 



(i) Page 105, line 4 If v = (v 1} v 2 , v 3 ) then the ith coordinate of Cv is 2q;vj, 

so that J 

(Cv) =22c i jVjU i (q) 

for any point q. In particular, putting q = F(p) gives 
(CV) F(P) = ?fijvjUi(F(p)). 



M334 III.2 15 

But F *( v p ) = ( Cv )f( p )' and so 

^(v p ) = SZcyVjUifFfc)) = 2 c ijVj Ui(F(p)). 
1 J 1 »J 



F, 



However, v p = 2)v;Uj(p), so this becomes 

fJx 



SvjUjCp)! = i 2c i jv j U i (F(p)), 



which can be abbreviated to the result given in line 4 by omitting the "p"s 
and writing Ui for Uj(F). 



Summary 

Results 

(i) The derivative map of an isometry F = TC sends 

the tangent vector v at the point p to the tangent 
vector Cv at the point F(p). Page 104, Theorem 2.1 

(ii) If F is an isometry, F* p is an orthogonal trans- 
formation from T (E 3 ) to Tp, ,(E 3 ). Page 105, Corollary 2.2 

(iii) Given two frames in E 3 , there is a unique isometry 

taking one onto the other. Page 105, Theorem 2.3 

Techniques 

(i) Finding the derivative map of an isometry using 

Result (i). Page 104, Theorem 2.1 

(ii) Finding the isometry mentioned in Result (iii), 
using the method employed in the proof of 
Theorem 2.3. Page 105, line -3 of text 



Exercises 

Technique (i) 

1. Page 106, Exercise 1. 

Technique (ii) 

2. Page 107, Exercise 5. 

Theory Exercises (omit if short of time) 

3. Page 106, Exercise 2. Write F = T a A and G = T^B and use the results of 
Exercise III. 1.2, whose solution occurs on Text page 13, and Exercise 
III. 1.3, whose solution occurs on Text page 13. 

4. Page 107, Exercise 4(a). (HINT: r is in the plane through p orthogonal to q if 
and only if (r - p)«q = 0.) 



16 
Solutions 



M334 III.2 



Page 106, Exercise 1. 

We can write T = TI for any isometry T, where I is the identity mapping. In 
the case when T is itself a translation this expresses T in standard form, as I 
is an orthogonal transformation. Using Theorem III.2.1 we have 



T *( v ) = ( Iv )t( P ) 



'T(p)' 



because Iv = v. 



Thus T*(v) has the same Euclidean coordinates as v; only the point of 
application has changed. In other words, T s|e (v) is parallel to v. 

Page 107, Exercise 5. 

The attitude matrix of the e frame is 



u 



A = 



and the attitude matrix of the f frame is 
/ 1 



B = 



V2 


1 
V2 







By the argument at the end of the text on page 106 we must have that 
/ 1 



C = X BA = 



V2 



V2 

1_ 

V2 

2 
3 







- 



V2 


_J_ 

V2 

72" 

2 
3 



V2 V2 V2, 



W 



M334 III.2 

and T is the translation through 



17 



q-c( P ) = 



/ 



-l 



_2 
3 







V2 

2 
3 



^ \372 372 -V2/\°/ 



-i 



w 



= 


4 
"3 










W 






Thus the isometry F is given by 






/M/* 





1 
V2 


F(r) = 


4 
3 


+ 


2 
3 


1 
3 


2 
3 


1 


1-^ 2 J 




1 


4 


1 



\3y/2 V2 V2^ 



for all points r = (r 1} r^ r 3 ) of E 3 . 

Let us check that F really does carry the e frame into the f frame; that is, 
F^ejJ = fj for i = 1, 2, 3;that is, F(p) = q and Cej = fj for i = 1, 2, 3. 



F(p) = 



1- 



2v/2 



fr 





V2 


2 
3 


1 
3 


2 
3 


1 


4 


1 


\V2 


V2 


V2 



18 



M334 III.2 



/• \ I 



1- 



2>/2 



I »\ 



\V2/ 



-1 



\ ' 



Ce! = 



V2 

2 
3 



1 
V2 

2 
3 



V2 V2 V2 



1 
V2 



Ce,= 



V2 V2 V2 



Ce,= 



V2 

_2 
3 




V2 V2 V2 



Page 106, Exercise 2. 

If F = T a A and G = T^B, where A, B are the orthogonal parts of F and G, 
then, by the result of Exercise III. 1.2, GF has orthogonal part BA. Hence, by 
Theorem III.2.1, if v p is any tangent vector, 

(GFMv p ) = (BAv) GF(p) . 

The same theorem gives 

F*(v p ) = Av F(p) 

and, applied to the isometry G and the tangent vector F^v-), 
G*(F*(v p )) = B(Av) G(F(p)) , 



M334 III.2 19 

that is 

(G*Fj(v p ) = (BAv) GF(p) . 

Hence (GF)*(v p ) = (G^F^Vp) for all tangent vectors v p , and so it 
follows that (GFJ* = G^F*. 

By the result of Exercise III.1.3, F" 1 has orthogonal part A" 1 , and so 
Theorem III. 2.1 gives 

(F-')»(v p ) - (A-v) Fl(p) 

for any tangent vector v p . We must show that this is the same as (F*)" 1 (v p ). 
We know that 

F*(v p ) = ( Av ) F(p) > 

that is, F* acts on the vector part of v« by the orthogonal transformation A 
and on the point of application of v p by F. Thus (F*)" 1 , which is the inverse 
of F*, must act on the vector part of v p by A" 1 and on the point of 
application of v p by F" 1 ; that is, 

(FJ- 1 ^) = (A-M F -i (p) - 

Consequently 

(F*)- 1 ^) = (F-%(v p ) 
for all v p , and so (F*)- 1 ^- 1 )*. 
4. Page 107, Exercise 4(a). 

r is in the plane through p orthogonal to q if and only if 

(r-p).q = 0. 
C preserves dot products so this happens if and only if 

(C(r- P )).C(q) = 0. 

Suppose T is translation by the vector a. 

C is linear, so 

C(r-p) = C(r)-C(p) 

= C(r) + a-(C(p) + a), 

= F(r)-F( P ). 
Thus 

(C(r-p)).C(q) = (F(r)-F(p)).C(q), 

which is zero if and only if F(r) is in the plane through F(p) orthogonal to 
C(q). Hence F carries the plane through p orthogonal to q to the plane 
through F(p) orthogonal to C(q). 



20 



M334 III.3 



III.3 ORIENTATION 



Introduction 



This section continues from Section III. 2, also tying in ideas from Unit M201 24, 
Orthogonal and Symmetric Transformations, Sections 2.2 to 2.4. You may recall 
from that unit that frames in E 3 are of two distinct types, known as left-handed and 
right-handed, distinguished by which hand "best fits" the frames. Moreover, ortho- 
gonal transformations of E 3 also fall into two distinct sets: there are the proper 
rotations (that is, those transformations physically realizable in 3-space as a rotation 
about an axis) and the improper rotations (that is, those transformations which are 
not realizable as an actual movement in 3-space because they consist of a proper 
rotation followed by a reflection). It turns out that proper rotations preserve the 
"handedness" of frames, whereas improper rotations interchange left-handed and 
right-handed frames, and that the corresponding matrices have determinant +1 for a 
proper rotation and - 1 for an improper rotation. 

In this section we give a more mathematical definition of "right-handed" and "left- 
handed" and classify all isometries of E 3 (not just orthogonal transformations) in 
terms of their effect upon frames. We are able to state and prove the results quite 
precisely, in a form that makes them suitable for use in proving the important 
theorems of Section III.5. 

READ: Sections III. 3 (pages 107-110). 



Comment 

(i) Page 107: Remark 3.1 

(1) In this case the attitude matrix is I, and det I = +1. 

(2) Since e 3 is a unit vector orthogonal to both ej and e 2 , it must be 
either e!Xe 2 or -(e^ej). By the result of Exercise 11.1.4(d), 
ei*e 2 X e 3 = e 3 '(e 1 Xe 2 ), which is +1 if e 3 = e! X e 2 , -1 if e 3 = ~(c\ X e 2 ). 

(3) This is a restatement of part of Lemma II. 1. A, on Text, Part II, 
page 7. 



Supplementary Comments 

(i) Page 110: the determinant Remember that a formal expansion of 



yields 



ei 


e 2 


e 3 


Vl 


v 2 


v 3 


w x 


w 2 


w 3 



(v 2 w 3 - v 3 w 2 )e! + (v 3 W! - V!W 3 )e 2 + (vjw 2 - v 2 Wi)e 3 . 



M334 III.3 21 

(ii) Page 110: proof of Lemma 3.5 If the frame is negatively oriented, the 
ex component of v X w is 

v 2 e 2 X w 3 e 3 + v 3 e 3 X w 2 e 2 = (v 2 w 3 - v 3 w 2 )e 2 X e 3 

= (v 2 w 3 - v 3 w 2 )(-e!), by Remark 3.1(3), 

= e (v 2 w 3 - v 3 w 2 )e! 

because e = - 1 when the frame is negatively oriented. 



(iii) Page 110: end of proof of Theorem 3.6 



F*(v) X F*(w) = e 



ei 


e 2 


«s 


Vl 


v 2 


v 3 


Wj 


w 2 


w 3 



= e ((v 2 w 3 - v 3 w 2 )e! + . . .) 

= e ((v 2 w 3 - v^F^Uifr)) + • • •) 

= e (F 5lc ((v 2 w 3 - v^U^p) + . . .)), 



= e 



/ 



\ 



U^p) U 2 (p) U 3 (p) 
vi v 2 v 3 

Wi w 2 w 3 



1 



= e F^v X w), 



and 



e = ei .e 2 X e 3 = F :)c (U 1 (p)).F 5!c (U 2 (p)) X F*(U 3 (p)) 
= (sgnF)U 1 (p).U 2 (p)XU 3 (p), 



by Lemma III.3.5, 



because F% is 
linear, 



by the definition 
of cross product, 



by Lemma III.3.2, 



= sgnF, 



because the frame 
U!( P ), U 2 (p), 
U 3 (p) is posi- 
tively oriented. 



Summary 



Notation 
sgnF 



Page 108, line 8 



Definitions 
(i) 



(") 



Positively oriented ) 
Right-handed f 

Negatively oriented \ 
Left-handed J 



Page 107, line -17 
Page 107, line -16 



22 M334 III.3 

(iii) Sign of an isometry, sgn F Page 108, line 6 

(iv) Orientation-preserving Page 109, Definition 3.3 

(v) Orientation-reversing Page 109, Definition 3.3 

Results 

(i) The natural frame field is positively oriented. Page 107, Remark 3.1(1) 

(ii) The frame e 1? e 2 , e 3 is positively oriented if and 

only if e 2 X e 2 = e 3 . Page 107, Remark 3.1(2) 

(iii) Frenet frames are positively oriented. Page 107, Remark 3.1(2) 

(iv) If e u e 2 , e 3 is a positively oriented frame, 

e[ = e; X e^ = - e^ X e; 

for (i, j, k) = (1, 2, 3) or (2, 3, 1) or (3, 1, 2). Page 107, Remark 3.1(3) 

(v) F^(e 1 )-F ! ,(e 2 ) X F*(e 3 ) = sgn F e r e 2 X e 3 . Page 108, Lemma 3.2 

(vi) The coordinates of tangent vectors in terms of a 
positively oriented frame give the expected for- 
mula for the cross product; if the frame is nega- 
tively oriented the usual formula must be multi- 
plied by - 1. Page 110, Lemma 3.5 

(vii) Orientation-preserving isometries preserve cross 
products; orientation-reversing isometries change 
their sign: 

F^v X w) = sgn F F^v) X F*(w). Page 110, Theorem 3.6 

Techniques 

(i) Distinguishing positively and negatively oriented 

frames. Page 107, lines -17, -16 

(ii) Finding the sign of an isometry Page 108, lines 6 to 8 

(iii) Finding the effect of an isometry on a cross pro- 
duct, using Result (vii). Page 110, Theorem 3.6 



Exercises 

Technique (i) 

1. Find the orientation of each of the frames in Exercise III. 2. 5 on page 107. 

Technique (ii) 

2. Find the sign of the isometry in Exercise III.2.5 and check that Lemma 
III.3.2 is satisfied. 

Technique (iii) 

3. Page 111, Exercise 3. 



M334 III.3 
Theory Exercises 



23 



(omit if short of time) 



4. Page 111, Exercise 1. (HINT: Use the result of Exercise III. 1.2.) 

5. Page 111, Exercise 2. 



Solutions 

1. The attitude matrix of the e frame is 

2 2 1 

3 3 3 



111 
3 3 3 

\l _2 2 / 

\3 3 3/ 



whose determinant is 

2|l.2_2j_2\\ + 2/2a_(_42\ + l/_2/ 2\ 1.1 1 
3\3 3 3 \ 3J] 3\3 3 \ 3/ 3/ 3\ 3 \ 3/ 3 3 

= -^(2(2 + 4) + 2(2 + 4) +(4-1)) 

= 1, 
so the e frame is positively oriented. Alternatively (and more simply) 

e,Xe 2 = ±(2.2 - 1.1, 1.-2 - 2.2, 2.1 - 2.(-2)) 
= I(3,-6,6) = e 3 
and so the e frame is positively oriented by Remark 3.1(2). Similarly, 
fi* h =-72(0.0- 1.1, 1.0- 1.0, 1.1- 0.0) 



V2 



(-l,0,l)=-f 3 



and so the f frame is negatively oriented. 

Lemma III.3.2 is satisfied if F 5N (e 1 )»F J|c (e 2 ) X F*(e 3 ) = sgn F e^ X e 3 . We 
found in the previous section that 



1 



C = 



1 

V2 

2 
3 

V2 V2 '372j 



V2 

_2 
3 







24 M334 III.3 

sgn F . det c = _>_(|. _ _^ . 1. _y + + j- 2 [l-fc - L-fe 

=^ei-8-8-i)=-i. 

Since sgn F = - 1, Lemma III.3.2 predicts that the e frame and the F(e) frame 
(that is the f frame) have opposite orientations: we have already shown that 
this is the case in the previous exercise. 

3. Page 111, Exercise 3. 

We want to show that C*(v X w) = sgn C C*(v) X C^w). 

v = (3, 1,-1) 

w = (-3,-3, 1) 
so 

v X w = (1 - 3, 3 - 3, -9 + 3) = (-2, 0, -6). 
C is its own orthogonal part, so C* is represented by the same matrix as C. 



C*(v) = 



4 -i 



C*(w) = 




1 

3 

11 
3 

7_ 
"3 



C*(vX w)= 




C*(v) X C*(w) = (- 1, 3, 1) X |-i -^, -|] 



lf\ 



8_ 
3 

14 
3 



, 10 8 14, n . v . 



so the formula is checked if sgn C = - 1. 



M334 III.3 25 

However, 

2 1 _ 2 2\_ l/2.2 _ 1.1 

3 3 3 3 3 3 3 3 3 



sgnC = detC=-|[|-| + |-|)+f(- 

= ^(-2(2 + 4) + 2(-2-4)-(4-l)) 



= -1 

and so the formula is checked. 

4. Page 111, Exercise 1. 

Let F = T a A, G = T^B. Then, by the result of Exercise III.1.2, FG has ortho- 
gonal part AB. Hence 

sgn (FG) = det (AB), by definition, 

= (det A) (det B) 

= (sgn F)(sgn G), by definition of sgn F, sgn G. 

Similarly 

sgn (GF) = (sgn G)(sgn F) 

= (sgn F)(sgn G). 

Putting G = F" 1 , gives 

sgn (FF" 1 ) = (sgn F)(sgn F' 1 ). 

But FF" 1 = I, the identity mapping, and so 

(sgn F)(sgn F" 1 ) = sgn I = +1 : 

therefore 

-1- 1 



sgn F" 1 = 



sgnF 
= sgn F, because sgn F = ±1. 

Page 111, Exercise 2. 

Let G be an orientation-reversing isometry of E 3 . We want to express G as 
HqF, so we define F to be Ho _1 G. Then certainly 

H F = H Hq^G = G 

and so all that remains to be shown is that F is orientation-preserving. Now, 
by the result of the previous exercise, 

sgn G = sgn(H F) = (sgn H )(sgn F). 
Here G and Hq are both orientation-reversing, and so 

sgnG = sgnH = -l; 
hence sgn F = +1 and so F is orientation-preserving. 



26 M334 III.4 

III.4 EUCLIDEAN GEOMETRY 

Introduction 

This section depends on Sections III. 1 -III. 3 and Section II. 3. Now we are able to 
use the ideas and results of Sections III. 1 - III. 3 to begin to define what we mean by 
"the same, but for position in space, as . . .". Loosely speaking, we are going to 
declare two "objects" in E 3 (for example, curves or vector fields) to be "the same 
but for position in space" if there is an isometry carrying one into the other. In 
order that this definition agree with our intuitive ideas of "sameness", two con- 
ditions must hold: (a) the relation "is the same, but for position in space as" must 
be an equivalence relation (any sensible definition of "same" must lead to an 
equivalence relation); (b) any properties of the objects concerned that we regard as 
essential, such as the curvature of a curve, the derivative of a vector field, must be 
preserved by isometries, so that objects which are "the same" have the same 
essential properties. Condition (a) is easily checked: the relation is transitive because 
the composite of isometries is itself an isometry (Lemma III. 1.3); it is reflexive 
because I is an isometry; it is symmetric because each isometry has an inverse iso- 
metry (Comment (i) on Section III. I). Condition (b) is the concern of this section, 
where it is shown that isometries preserve derivatives of vector fields on curves and 
the Frenet apparatus of unit-speed curves. 

READ: Section III.4 (pages 112-115). 



Additional Text 

(i) Arbitrary-Speed Curves and Isometries While it is true that any mapping 

F preserves the velocity of a curve a in the sense that 

F^cx') = (F(o0)' 

(Theorem 1.7.8), the speed of a curve is not generally preserved, because 
||a'|| and ||(F(a))'|| are not necessarily the same. However, in the case where 
F is an isometry an argument similar to that given at the beginning of the 
proof of Theorem 4.2 shows that F does preserve speed: F preserves norms, 
as noted on page 105, and so 

v F(o) = l|(F(«))'li 

= ||F.(«')H 
= Hall 



= V 



a* 



We have shown that a and F(a ) have the same speed if F is an isometry. The 
following exercise shows that they have the same curvature, and plus-or- 
minus the same torsion. 

Let a be an arbitrary-speed regular curve in E 3 , F an isometry of E 3 , 
= F(a), and a a unit-speed reparametrization of a, where a = a(s). 
Prove that K Q = K a and T Q = sgn F r a . (HINT: Use the above result to 
prove that j3 - F(a) is a unit-speed reparametrization of j3 in the sense of 
Theorem II. 2.1, and then use Theorem III.4.2.) 



M334 III.4 27 

Supplementary Comments 

(i) Page 113: second half As usual we are writing F = TC and C = (qj). 

(ii) Page 114: fourth line of proof T and T are defined by T = 0', 
T = j3 ', so the result follows using the fact that |3 ' = (F(j3))' = F*(j3') by 
Theorem 1.7.8. 



(iii) Page 114: seventh line of proof K and k are defined by K = ||jS"||, 
K=p"||; j3"=F*(0") by Corollary HI.4.1; ||F*(j3")|| = ||0" \\ because 
F % preserves norms. 

(iv) Page 114: eleventh line of proof (¥^"))/k = F*(j3"/k) because F^ 

is a linear transformation at each point. 



Summary 

Results 

(i) Isometries preserve derivatives of vector fields on 

curves: (F*(Y))' = F*(Y'). 

In particular, isometries preserve acceleration. Page 113, Corollary 4.1 



(ii) Isometries preserve the speed of curves. Text, Page 26 

(iii) Isometries preserve the Frenet apparatus of unit- 
speed curves, except that orientation-reversing iso- 
metries change the sign of the torsion and reverse 
the direction of the binormal: 

k = k T=F !tc (T) 

r = sgnFT N=F*(N) 

B = sgn F F !|C (B) Page 114, Theorem 4.2 

(iv) A similar result for arbitrary-speed curves: if 

j8 = F(a), then Kq = K a and r Q = sgn F r a . Text, Page 26 



Techniques 

(i) Applying Result (i) to find the derivatives of vector 

fields of the form F 5|C (Y) on a curve F(a). Page 113, Corollary 4.1 

(ii) Using Result (iii) to find the Frenet apparatus of 

curves of the form F(a). Page 114, Theorem 4.2 



28 
Exercises 



M334 III.4 



Technique (i) 

2. Page 115, Exercise 2. (Remember that, because C is orthogonal, C* has the 
same matrix as C.) 

Technique (ii) 

3. Page 115, Exercise 1. 



Solutions 



We have seen that a and |3 have the same speed, and consequently a. and /3 
have the same arc-length function s. Now 

?(s) = (F(ff))(s) = F(n(s)) = F(a) = j8 

and so j3 is a unit-speed reparametrization of )3 in the sense of Theorem 
II.2.1. 

Hence 

by the definition of the curvature and torsion of an arbitrary-speed curve. 
Similarly 

K a =K S (s) r a =7ff(s). 

However, ZX is unit-speed and j8 = F(S) and so Theorem III.4.2. gives 

r^=sgnFr s -. 



K.Q K s 



Hence 



and 



K P = K P^ = K ^ = Kql 



T |8 = T /^ S ) = Sgn F T ^ S ^ = Sgn F T «* 
Page 115, Exercise 2. 

- 1 cost 



a(t) = C(cx(t)) = 







\/2 V2 



1 
V2 



V2 



sin t 



2t 



-cost 



-^(sint- 2-t) 
,-^(sint + 2t) 



/ 



M334 III.4 



29 



Y(t) = C*(Y(t)): since C is itself an orthogonal transformation C* has the 
same matrix as C and so 



l- 







C*(Y(t)) = 



\ 



" V2 "72 

\° 72 72/ 



1-t 2 



\ 1+t2 / 



V2t 2 
V2 



That is, 



Thus 



Now 



and so 



Y(t) = (-t,V2tV2). 



Y'(t) = (-l,-2V2t,0). 



Y'(t) = (l,-2t,2t) 



t- 



C*(Y'(t)) = 





1 1_ 

72 "V2 

1 1 
\/2 V2 



-2t 



2t 



-2V2t 



Thus Y'(t) = C*(Y'(t)) for all t, and so Y' = C*(Y'). 
a'(t) = (-sint, cost, 2), 
a"(t) = (-cos t, -sin t, 0) 



30 



so 



/- 



C*(a"(t)) = 



1 

V2 

1 

V2 



1 
V2 

1 

V2 



/ -cost 1 



-sint 



cost 



V2 

1 

V2 



suit 



sint 



whereas 



M334 III.4 



a'(t) = sint,-i(cost - 2), -^(cost + 2) 

— «/,v I . sint sint' 
a(t) = |cost, --^ --^ 

and so C*(a"(t)) =a"(t) for all t; that is, C*(a") = 2e". 
We have already computed Y', a ", Y , a": from this we see that 
(Y'V)(t)= (l,"2t, 2t)-(-cost, -sint, 0) 
= -cost + 2tsint 



and 



so 



(Y'.«")(t) = (- 1, -2V2t, 0)- cos t, 



= -cost + 2t sint 



Y •« = Y -a . 



-sint -sint 
V2' V2 



3. Page 115, Exercise 1. 

(a) Since |3 has unit speed, T = ]3', and so if /3 is a cylindrical helix then there is a 

constant unit vector u and a constant angle # such j3'«u = cos #. Let 
ii = F^u). Then 

H»|| = ||f,<u)H = Hi = i 

and so u is a constant unit vector. Moreover, F(j3) also has unit speed; that is, 
T = F(|8)', and so 

T-u = (F(0))'.u 

= (F(0))'.F» 

= F*(/3>F*(u), 

= COS #, 

and so F(j3) is also a cylindrical helix. 



by Theorem 1.7.8, 

because F ^preserves dot products, 



M334III.4 31 

(b) Let F(j3) denote the spherical image of F(j3), T the unit tangent vector field 
on F(j3). Then F(p) is the curve whose points have coordinates given by T, 
while ft is the curve whose points have coordinates given by T. In other 
words 

T(t)=|3(t)0 (t) and T(t) = F$)(t) F (/?(t))- 
Omitting the t: 

T = J?0 and T=F03) F(i3 ). 
By Theorem III.4.2, T = F^(T) 
andsoF(jS)F^) = F Hc (|3|3). 
Since F = TC, Theorem III. 2.1 tells us that 

F*(00) = CflJ ) F 0) 

and so 

^)F(P) = C(?) F( ^ 

and so 

F(0) = C(/3). 



32 M334 III.5 

ffl.5 CONGRUENCE OF CURVES 

Introduction 

At last we reach the definition of congruence of curves promised in Section III.l 
This is the definition suggested in Section III. 4 and, as the remarks there show, 
congruence is in fact an equivalence relation. Results from all the earlier sections of 
Chapter III are used to prove the main theorem that unit-speed curves are congruent 
if and only if their curvatures are the same and their torsions differ at most in sign. 
Using the definitions of Section II.4 we can widen this theorem to include arbitrary- 
speed regular curves. Finally we generalize the main theorem to non-Frenet frame 
fields, using the ideas of Section II.6. This last theorem (Theorem III.5.7) is less 
important now but will be needed in Chapter VI: you are not expected to remember 
its details. 

Some of the proofs in this section are a little long and technical. Try to work 
through them, but if you get bogged down make sure that you at least understand 
the statements of the theorems: they are the whole purpose of this chapter. 

READ: Section III.5 (pages 116- 121). 



Comments 

(i) Pages 117-118: proof of Theorem 5.3 Notice the structure of the 

proof. If F is to carry a to j3 then certainly F(a(t)) = |3(t) for all t, so our first 
requirement is that F(a (0)) = 0(0). Secondly, we know from Theorem 
III.4.2 that F* must take the Frenet frame of a at a (0) to the Frenet of j3 at 
0(0), reversing the direction of the binormal if r a =-r^, so we require that 
F* have this effect on T a (0), N a (0), B a (0). These two conditions specify F 
uniquely, so the final stage is to check that F(a ) is actually equal to j3. 

Notice also the beauty and economy of the last stage of the proof. We know 
that T«T<1 because 

T-T = |ffl|||T||cost>, 

where # is the angle between T and T, and cos # = 1 if and only if # is a 
multiple of 2ir, that is if and only if T and T are equal. Thus we can show 
that T = T by showing that T-T is always 1. However, because the same 
remarks apply to N and N, B and B, we can show that all three pairs are 
equal by the single step of showing that 

f = T-T + N-N + B-B= 3. 

This is shown by a technique we have noted before, that of proving f' = to 
show f is constant and noting the f(0) has the correct value. 

(ii) Page 119: Corollary 5.5 When O'Neill uses the word "helix" 

unqualified, he means not a general cylindrical helix, but a circular helix, 
that is a cylindrical helix whose cross-section curve is (part of) a circle (see 
Exercises II.4.8 and II.4.10), as he remarks in Example 1.4.2(2). An equiva- 
lent definition of circular helix is as a curve congruent to a reparametrization 
of the helix 

1 1 — ► (a cos t, a sin t, bt) 
given in Example 1.4.2(2), and it is this definition O'Neill uses in the proof 
here. 



M334 III.5 33 

Supplementary Comments 

(i) Page 117: line -9 F is orientation-preserving because it carries one 

Frenet frame into another, and Frenet frames are always positively oriented. 

(ii) Page 117: line -2 

k = k. w by Theorem III.4.2, 

= Kg, by the hypotheses of the theorem. 

(iii) Page 119: Example 5.4 The "a" used here is simply a shorthand for 
l/\/2 = 1/c: it is not the same "a" as in Example II.3.3. 

(iv) Page 120: first two lines of proof a and |3 must be based at the same 
value of t so that a. and |3 have the same arc-length function s, for if 6t is 
based at tj and ]8 at t 2 then 

s a (0 = \ v a ( u ) du 
J U 

= \ Vg(u)du, because v a = v„, 

= \ V u ^ du+ \ V u ^ du 

= s + S/3 (t), 
and the constant s~ is zero if and only if t x = t 2 . 

(v) Page 121: Theorem 5. 7, hypotheses (1) and (2) These are just natural 
extensions of the hypotheses in Corollary 5.6, for if {Ej} and {Fj} are the 
Frenet frame fields on a, j3 respectively then condition (1) becomes 
simply v a = v~ and condition (2) becomes K a = k„, T a = t«. 

(vi) Page 121: lines 7; 8 of proof Since F carries the Ej frame at a (0) to the 
Fj frame at j3(0), the two left-hand equations are straightforward. For the 
other two: 

a'«Ej = (F(a ))'«F !( .(Ej), by definition, 

= F^a )«F^(Ej), because F,,. preserves velocities, 

= a -Ej, because F^ preserves dot products, 

= j3'*Fj, by hypothesis (1); 



also 



E}-Ej = (F 5lc (E i ))'.F :JC (Ej), by definition, 

= F 5|C (E-)'F % (E-), because F* preserves derivatives of 

vector fields, 

= Ej «Ej, because .F* preserves dot products, 

= Fj'«Fj, by hypothesis (2). 



34 M334 III.5 

(vii) Page 121: line 9 of proof Since {Ej} is a frame, Ej' has an orthonormal 
expansion Ej = 2 ajjEj, where aj; = Ej'«Ej (Theorem II. 1.5). Similarly, 

Fj' = 2 bjjFj, where by = Fj'-Fj. We have just shown that Fj'-Fj = Ej'*Ej> so 

D ij ^ a ij and we can write Fj' = 2 ajjF;. 



(viii) Page 121: lines 10, 11 of proof Ej«Ej = 5jj =^(Ej'Ej)' = 

==>Ej'.Ej + Ej.Ej' = 0= 



•a ij + aji = 0. 



(ix) Page 121: line 13 of proof 

f = 2(Ej'.F i + Ej.Fj') 
i 

= 2Ej'.Fj + SE;.Fj' 
j j j j 

= .2(aijEj-Fj + a ji E j .F i ) 
1 > J 

= 2(a i j + ajj)Ej.F i . 
lj J 

(x) Pag-e i2i: /me i4 of proof For i = 1, 2, 3, since_Ej(t) and Fj(t) are unit 
vectors, Ej(t)-Fj(t) < 1 with equality if and only if Ej(t) and Fj(t) are equal. 
Thus f(t) = 2 Ej(t)-Fj(t) < 3 with equality if and only if each Ej(t) is equal 

to the corresponding Fj(t). We have shown that f' is the zero function, so f 
itself is a constant function. However, f(0) = 3 (because Ej(0) = Fj(0) for 
each i) and so f(t) = 3 for all t, that is Ej(t) is equal to Fj(t) for all t. 



Summary 



Definitions 

(i) Congruent curves 

(ii) Parallel curves 



Page 116, Definition 5.1 
Page 116, line -1 



Results 
(i) 



If a '(s) and j3'(s) are parallel for each s, a and j3 are 

parallel. Page 117, Lemma 5.2 



(ii) If a and j3 are parallel and a(s Q ) = 0(s o ) for some 
s^, then a = ]3. 

(iii) Unit-speed curves a, |3 are congruent if and only if 
K a = Kp and T a = ±Tp 

(iv) Arbitrary-speed regular curves a, ]3 are congruent if 
and only if 



Page 117, Lemma 5.2 

Page 117, Theorem 5.3 and 
Page 114, Theorem 4.2 



Page 120, Corollary 5.6 and 
Text, page 26 



M334 IH.5 35 

(v) If a, |3 are arbitrary curves and {Ej}, {Fj} are 
frame fields on each such that 

a'-Ei = j3'-Fi 

and Ej'«E; = Fj'-F;, 

then a and j3 are congruent. Page 121, Theorem 5.7 

Techniques 

(i) Recognition of congruent curves. Page 116, Definition 5.1 

(ii) Finding an isometry establishing the congruence of 

two curves. Page 117, proof of 

Theorem 5.3 

Exercises 

Techniques (i) and (ii) 

1. Page 122, Exercise 6. 

Theory Exercise 

2. Page 122, Exercise 5. 

Solutions 

1. Page 122, Exercise 6. 

a(t) = (v/2t, t 2 , 0) 
a'(t) = (v/2, 2t, 0) 
v«(t) = V2(l + 21 2 ) 2 " 
a"(t) = (0, 2, 0) 
We use the formulas of Theorem II.4.3. 
a'(t) X a"(t) = (0, 0, 2V2) 



so 



so 



2>/2 1 

Ka(t)_ 2 v / 2(l + 2t 2 f (l + 2tf 

a'"(t) = so r a (t) = 0. 
|3(t) = (-t,t,t 2 ) 
/3'(t) = (-l, 1, 2t) 

v^(t) = (2 + 4t 2 ) i =V2(l + 2t 2 ) i = v a (t) 

0"(t) = (0, 0, 2) 

0'(t) X 0"(t) = (2, 2, 0) 

2y/2 1 

K {t) ' 2V2(1 + 2t 2 )f " (1 + 2t 2 )T " Ka(t) - 

/3"'(t) = so T0(t) = O = r a (t). 



36 M334 HI.5 

Thus <x and /3 have the same speed, curvature and torsion and so they are 
congruent. 

a'(0) = (v/2, 0, 0) so T a (0) = (1, 0, 0); 

a'(0) X a"(0) = (0, 0, 2^2) so B a (0) = (0, 0, 1); 

N a (0) = B a (0) X T a (0) = (0, 1, 0): 

0'(O) = (-1,1,0) so Tp(0) = l--^,±0\; 

0'(O) X /3"(0) = (2, 2, 0) so B0(O) =|^,^,o); 

N^(0) = B^O) X T^(0) = (0,0,1). 

Since r a = ra we want to find the isometry F carrying the a Frenet frame to 
the j3 Frenet frame: such an isometry has orthogonal part with matrix 



C = 



1 
V2 

1 

V2 



1 
V2 



V2 



V2 





1 

V2 


1 
V2 





1 

V2 



transpose of attitude 
matrix of |3 frame 



attitude matrix of (X 
frame 



Since a(0) =]3(0) = 0, C(a(0)) = |8(0) and so the translation part of the F is 
just the identity. 



Checking: 



1 
72 



72 



F(a(t)) = 



1 

V2 





1 

V2 





1 







=«t). 



Alternatively, since t ol = -tq also, we could choose F to take the a Frenet 
frame to the frame Tj3(0), N/j(0), -B«(0). This gives 



C = 



1 
V2 

1 

72 





1 
V2 

1 




01 [72 ° -72 



72 




1 



72 



and again the translation part of F is the identity. 



M334 III.5 37 

2. Page 122, Exercise 5. 

If a and are congruent, then K a = kq = k and T a = ± T0 = r. Suppose F 
is an isometry such that F(a) = )3. Then, as noted in the Additional Text (i), 
we must have 

F(«(0)) = «0) 

F !|e (T a (0)) = T^(0) 

F*(N a (0)) = N^(0) 

F*(B a (0)) = sgn F 1^(0) (*) 

by Theorem III.4.2. This theorem also tells us that 

T|3 = sgn F r a . 

If r # 0, this equation tells us that sgn F = 1 if T a = Tg, sgn F = - 1 if 
r a = -tq. Thus the four equations (*) determine the isometry F uniquely, 
and the proof of Theorem III.5.3 shows that in fact F(oc) = j3. 

If r = then there is no constraint on sgn F, so we can have sgn F = 1 or 
sgn F = -1. In each of these two cases the equations (*) give a unique iso- 
metry F. Moreover, since r a = tr and T a = - tq, the proof of 
Theorem III.5.3 shows that both of these isometries satisfy F(a) = |3. 



38 M334 III 

FURTHER EXERCISES AND SOLUTIONS 

Section III.l 

Technique Exercises 

Page 103, Exercises 5(b), (c). 

Solutions 

5(b). (V2,-5,V2) 
5(c). (- V2, 1, V2). 



Section III. 2 

Technique Exercise 

Page 107, Exercise 4(b). (HINT: (0, 1, 0) has the same length as (1, 0,-1)^2.) 

Other Recommended Exercise 

Page 106, Exercise 3. This proves the result given at the end of page 106 in an 
alternative way. 



Solutions 

4(b). There are many possible solutions, each of which is obtained by putting a 
particular value of ft in the following. 

sin ft 1 cos ft 

V2 72" "72" 

C = cos ft sin ft 

sin ft 1 cos ft / 

72" V2 x/2 / 

and T is translation through 

i j. sm ^ + _1 o _ cos # i + sin ft 1_ 

1 2^2 >/2 2 ,x 2V2 V2 

3(a). The translation part of F carries C(0) = to p, so is Tp. 

The orthogonal part of F carries Uj(0) to c v so has the coordinates of the ej 
as its columns, so it is A. 



M334 in 39 

3(b). The e-frame at p is taken to the natural frame at by 
(TpA-^ATp^TA^A. 

The natural frame at is taken to the f-frame at q by TqB" 1 . 

(TqB _1 )(TA(-p)A) takes the e-frame at p to the f-frame at q and has ortho- 
gonal part B"*A. 



Section III.3 

Recommended Exercises 

Page 111, Exercise 4. This shows that orthogonal transformations with determinant 
+1 can be realized as rotations of E 3 . 

Page 111, Exercise 6. Some more group theory. 

Page 111, Exercise 7. This looks at isometries of E 1 and E 2 , which are defined in 
the obvious way. 

Solutions 

4. Put e 3 = (p/||p||). Choose e! and e 2 to be orthogonal unit vectors spanning 

the subspace of E 3 orthogonal to e 3 . C(ex) is also orthogonal to e 3 and has 
unit length, so 

C(ei) = cos & e! + sin d e 2 
for some #. If 

C(e 2 ) = sin i? e! - cos # e 2 , 
replace & and e 2 by their negatives. 

6. For both parts: det (AB) = det A det B 

det A -1 = (det A) -1 
detl = +l. 

7. E 1 :F(x) = ex + a 

f cos & e sin ^ 

E 2 : F = TC, C = 



\ 



- sin ^ e cos # 

\ 

In each case, orientation-preserving if e = +1. 



Section III.4 

Technique Exercise 
Page 116, Exercise 3. 



40 
Solution 

3. 



M334 III 




T(3,-3) 



Section III.5 



Technique Exercises 



Page 122, Exercises 2,4. (HINT for Exercise 2: The Frenet apparatus is not needed 
— look for an isometry first.) 



Other Recommended Exercises 

Page 121, Exercise 1. This gives another characterization of congruence of curves. 

Page 122, Exercise 3. This uses Theorem III.5.7 and the connection forms. 

Page 122, Exercise 11. This proves a theorem equivalent to Theorem III.4.2 and 
Theorem III. 5. 3 for curves in the plane. 



M334 III 
Solutions 



41 



2. There are eight possible solutions. Each of e, 6, 17 may be ±1 in the follow- 

ing. 

7 (t) = (eV2t,53t 2 ,V2t 3 ) 



4. 



72 



F = 



* 



V2 



JL. 



K=ir=-i 








a=2,b = 2. 








/.. 


V3 
2 


1 
2 


F = 


1 










1. 


1 
2 


V3 

IT 



1. 

3. 



11. 



If F = T p C, where C(uj) = ej, then F(a) = 0. 

The frame field on a is (Ej(a)}, that on |5 is {Ej(0)}. 

a'.Ei(a)=0i( a '). 

(Ei(a))'.Ej(a) = (V a 'Ei).Ej(a) = «ij(a'). 
If: let F be the isometry taking 

a(0) to 0(0) 

T a (O)toT0(O) 

N a (0) to eN0(O) where kq = e/c a . 
Follow the proof of Theorem III.5.3. 
Only if: follow the proof of Theorem III.4.2. 



DIFFERENTIAL GEOMETRY 

I Calculus on Euclidean Space 

II Frame Fields 

III Euclidean Geometry 

IV Calculus on a Surface 

V Shape Operators/ 

VI Geometry of Surfaces in E 3 



M33' PART IV 



516. 

I 

DIF 



THE OPEN UNIVERSITY 
Mathematics: A Third Level Course 



9 



PART IV CALCULUS ON A SURFACE 



DIFFERENTIAL GEOMETRY 



9 



THE OPEN UNIVERSITY 

Mathematics: A Third Level Course 

DIFFERENTIAL GEOMETRY PART IV 



CALCULUS ON A SURFACE 



A commentary on Chapter IV of O'Neill's 

Elementary Differential Geometry 

Prepared by the Course Team 



THE OPEN UNIVERSITY PRESS 



Course Team 



M334 IV 



Chairman: 



Mr. P.E.D. Strain 
Dr. R.A. Bailey 



Lecturer in Mathematics 
Course Assistant in Mathematics 



With assistance from: 



Dr.J.M. Aldous 
Mr. GJ. Burt 
Dr. P.M. Clark 
Mr. P.B. Cox 
Dr. F.C. Holroyd 
Mr. T.C. Lister 
Mr. R.J. Margolis 
Dr. C.A. Rowley 
Mr. M.G. Simpson 



Senior Lecturer in Mathematics 
Lecturer in Educational Technology 
Lecturer in Physics 
Student Computing Service 
Lecturer in Mathematics 
Staff Tutor in Mathematics 
Staff Tutor in Mathematics 
Course Assistant in Mathematics 
Course Assistant in Mathematics 



Consultants: 



Prof. S.Robertson 



Prof. T. Willmore 



Professor of Pure Mathematics, 
University of Southampton 

Professor of Pure Mathematics, 
University of Durham 




The Open University Press, Walton Hall, Milton Keynes 

First published ^76vkeprinted 1980 

Copyright © 1976 The Open University 

All rights reserved. No part of this work may be reproduced in any form, by mimeograph or any other means, 
without permission in writing from the publishers. 

Produced in Great Britain By 
The Open University Pre/s 

ISBN 335 05703 9 

This text forms part of the correspondence element of an Open University Third Level Course. The complete list 
of parts in the course is given at the end of this text. 

For general availability of supporting material referred to in this text, please write to: Open University 
Educational Enterprises Limited, 12 Cofferidge Close, Stony Stratford, Milton Keynes, MK11 1BY, Great 
Britain. 

Further information on Open University courses may be obtained from The Admissions Office, The Open 
University, P.O. Box 48, Milton Keynes, MK7 6AB. 
1.2 



M334 IV 3 

CONTENTS 

Page 

Set Book 4 

Bibliography 4 

Conventions 4 

Introduction 5 

IV.l Surfaces in E 3 6 

IV.2 Patch Computations 20 

IV. 3 Differentiable Functions and Tangent Vectors 33 

IV.4 Differential Forms on a Surface 43 

IV.5 Mappings of Surfaces 49 

IV. 7 Orientability 56 

Further Exercises and Solutions 59 



4 M334 IV 

Set Book 

Barrett O'Neill: Elementary Differential Geometry (Academic Press, 1966). It is 
essential to have this book: the course is based on it and will not make sense without 
it. 



Bibliography 

I 

The set books for M201, M231 and MST 282 are referred to occasionally; they are 
useful but not essential. They are: 

D.L. Kreider, R.G. Kuller, D.R. Ostberg and F.W. Perkins: An Introduction to 
Linear Analysis (Addison- Wesley, 1966). 

E.D. Nering: Linear Algebra and Matrix Theory (John Wiley, 1970). 

M. Spivak: Calculus, paperback edition (W.A. Benjamin/ Addison-Wesley, 1973). 

R.C. Smith and P. Smith: Mechanics, SI Edition (John Wiley, 1972). 



Conventions 

Before starting work on this text, please read M334 Part Zero. Consult the Errata 
List and the Stop Press and make any necessary alterations for this chapter in the set 
book. 

Unreferenced pages and sections denote the set book. Otherwise 

O'Neill denotes the set book; 

Text denotes the correspondence text; 

KKOP denotes An Introduction to Linear Analysis by D.L. Kreider, R.G. 
Kuller, D.R. Ostberg and F.W. Perkins; 

Nering denotes Linear Algebra and Matrix Theory by E.D. Nering; 

Spivak denotes Calculus by M. Spivak; 

Smith denotes Mechanics by R.C. Smith and P.Smith. 

Reference to Open University Courses in Mathematics take the form: 
Unit Ml 00 22, Linear Algebra I 
Unit MST 281 10, Taylor Approximation 
Unit M201 16, Euclidean Spaces I: Inner Products 
Unit M231 2, Functions and Graphs 
Unit MST 282 1, Some Basic Tools. 



M334 IV 
INTRODUCTION 



This chapter uses the differential and vector calculus developed in Sections LI to 
ILL 

It introduces the second major theme of this course, surfaces. The surfaces in which 
we shall be interested are subsets of E 3 that locally look like some open region of 
E 2 . You have already met some of these, such as planes, spheres and cylinders. 

After discussing several ways of describing a surface we extend the definitions of 
tangent vector, differential form and mapping to apply to them. 

Finally we use differential forms to describe orientability of a surface. 



M334 IV. 1 



IV.l SURFACES IN E 3 



Introduction 



This section follows on from Sections 1.2, 1.4, 1.5, 1.7 and II. 1. 

We shall be using implicitly defined curves in the plane. You first met these in 
Section 1.4. Typical examples are: 




¥ x 




*x 



(a) C^ : y - x 2 = 



(b) C 2 :x 2 +y 2 =l 



In general a curve is defined as a subset C: g = c, where g is a differentiable function 
and c is a constant. We shall always assume that, for each point (x , y<>) in C, either 
(dg/3x) (x , y ) =£ or (3g/3y) (x , y ) =£ 0, and that if C has more than one 
component then we are restricting attention to just one of them. With these restric- 
tions the only curves that we obtain are similar to C^ and C 2 . 

(a) Given a curve like C t we can parametrize it by a regular differentiable 
function a from R to E 2 that is one-to-one and whose inverse map a -1 from 
C x to R is continuous. In our example of Cx such a parametrization is given 
by a(t) = (t, t 2 ). This has inverse or 1 (x, y) = x for points in C^ . 

(b) Given a curve like C 2 we can parametrize it by a periodic regular differenti- 
able function j3 from R to E 2 . We can also demand that if we restrict 
attention to small enough open subsets of C 2 they will be the one-to-one 
images of small intervals of R and the inverse functions between these small 
subsets will be continuous. In our example of C 2 such a parametrization is 
given by j8(t) = (cos t, sin t). A typical 'local' inverse for this function is given 
by IT 1 (x, y) = cos -1 x for all those points p in C 2 with y(p) > 0. 

We shall need the implicit function theorem, a two dimensional form of which was 
dealt with in an appendix to Unit M231, Applications of the Derivative, Section 8. 

As an example of an implicitly defined curve in E 2 we can consider again the unit 
circle 



C: f(x, y) = x 2 + y 2 = 1 



M334 IV. 1 




We can use the implicit function theorem to describe the local behaviour of this 
curve. The implicit function theorem states: 

Theorem IV. 1. A 

If the point (x , y ) belongs to C and if (3f/dy) (x , y ) =£ then there exists: 
(i) a neighbourhood U of x in the real line; 

(ii) a unique differentiable function g : U »R; and 

(iii) a neighbourhood N of (x , yo) in the plane; such that 
NnC={(t,g(t)):t€EU}. 



(x »y ) 




*x 



That is, locally the curve C can be described as the graph of a mapping g : U- 
Similarly we can describe C equally well in terms of y near (x , y ) if 
(df/dx)(x o ,yo)*0. 



-►R. 



8 



M334 IV. 1 



1 3 



In our example, if x = y we can choose for U the open interval (^-,-4) and then g is 
the function 

g:tl — ►(1-t 2 )*". 
For a suitably chosen N we have that 

NnC={(t,(l-1 2 f):{<t<|}; 

that is, N fl C is the graph of g. 

In this section we shall apply an extension of the implicit function theorem to 
implicitly defined subspaces of E 3 . For instance, near a point (x , y , zq) on the 
unit sphere 

X : f(x, y, z) = x 2 + y 2 + z 2 = 1 
for which 

(9f/3z) (x , y , z ) = 2z 

is nonzero, the sphere can be described as the graph of a function g, where 
g(x, y) = z = (l- x 2 - y 2 ) T . 



(x ,y , (l-x 2 -y 2 r) 




2\T\ 



( x o> yo»'0) 



READ: Section IV.l (pages 124- 131). 



Comments 

(i) The plane E 2 When we use E 2 , or some open set D in F^ 2 , as the domain 

for a patch x we write u and v for the natural coordinate functions rather 
than x and y or x l and x 2 . Consequently when we want to refer to a variable 
point in E 2 we write (u, v), though strictly speaking this is the identity 
function from E 2 to itself. Whichever interpretation you adopt it is unlikely 
that any ambiguity will arise. When we have to use the partial derivatives of a 
function f whose domain is in E 2 we write them as 8f/3u and 3f/8v. 



M334IV.1 



Page 124: Proper patches The conditions in the definition of a proper 
patch are there to rule out certain pathological possibilities. The condition 
that the mapping x is regular means that its derivative x^ is one-to-one and 
hence no nonzero tangent vector is mapped to a zero tangent vector. Since 
derivatives preserve velocities, each curve with nonzero velocity will be 
mapped to another curve with nonzero velocity. As we saw in Section 1.4 
this means that x does not introduce any kinks into the patch. The image of 
a patch cannot look like the following diagram. 




The condition that x is one-to-one rules out the possibility that the patch 
might intersect itself, as in the following diagram. 




The condition that x" 1 is continuous rules out the possibility that the patch 
might approach itself arbitrarily closely, as in the following diagram. 




10 M334 IV.l 

(iii) Page 125: Definition 1.2 To prove that a subset of E 3 is not a surface 
we need to find only one point that does not have any neighbourhoods that 
are like smooth open discs. One way to do this is to look at the boundaries 
of suitably small neighbourhoods to see if they are like circles. For instance 
in the second example of Comment (ii) the boundary of a typical small 
neighbourhood of a point on the line of self-intersection is shown in the 
following diagram. 




We shall assume that our intuition does enable us to distinguish this from a 
circle. 

(iv) Page 129: Curves and surfaces The curves used in both Example 1.5 and 
Example 1.6 are defined implicitly. As in Section 1.4 this rules out the 
possibility that they may be of any of the following forms. 





(a) 



(b) 



(c) 



This in turn ensures that the surfaces derived from them, as in Examples 1.5 
and 1.6, are not pathological in any of the ways mentioned in Comment (ii). 
In fact the implicit function theorem discussed in the introduction proves 
that the above types of curve cannot occur. Curves (b) and (c) cannot be 
expressed locally as graphs and curve (a) cannot be expressed as the graph of 
a differentiable function. 

When dealing with implicitly defined curves we shall always restrict ourselves 
to one component at a time. In Example 1.6 this enables us to assume that 
the curve is in the upper half plane. If we did not have this restriction we 
would have to consider curves like the following hyperbola. 



M334 IV.l 



11 




When this is revolved about the x axis the spaces traced out by the two 
components intersect and the result is not a surface. 

As mentioned in the introduction this restriction limits us to two types of 
curve. The following are typical examples. We can revolve them about any 
line that does not intersect them. 




► x 



(a) Ci : y - x 




► x 



For curves like C^ we shall use only one-to-one parametrizations and with 
curves like C 2 we shall use only periodic parametrizations. As usual all 
parametrizations of implicitly defined curves will be regular. 



12 



M334 IV.l 



Supplementary Comments 

(i) Page 129: Example 1.6 

ordinate fixed. 



Rotating about the x axis we keep the x co- 




>s *( ( li> <b cos v > 92 sm v ) 



/(qi»q2,o) 



Hence q 2 gives the radius of the circle that the point (qj , q 2 , 0) traces out as 
it is revolved. When it has been revolved through an angle v its new coordi- 
nates are (qj , q 2 cos v, q 2 sin v). 

If the point p = (pi, p 2 , p 3 ) is revolved about the x axis, its x coordinate 
remains unchanged and its distance from the x axis remain unchanged. This 
distance is (p 2 + p|) . Hence when it has been swung round into the upper 
half of the xy plane it reaches the point p = (pi , (p 2 + pi) , 0). The point p 
is on the surface of revolution if and only if this point is on the original 
curve. So (p! , p 2 , p 3 ) belongs to M if and only if f(p 2 , (p 2 + pi ) ) = c. That 

is, M is given by M : g = c where g(x, y, z) = f(x, (y 2 + z 2 ) ). 

How do we use the chain rule to prove that dg is never zero? Firstly, since g 

is a mapping from E 3 to R 



a-g(v) = v[g] =±(g(p + tv)) 



t = = g *( v ) 



for any tangent vector v to E 3 at p. Similarly for any tangent vector w to E 2 
at q then 

df(w) = f*(w), 

where f is a function from E 2 to R and the differential df is defined by 
analogy with the definition for mappings from E 3 to R. Now g is the com- 
posite mapping f<>F where 



F(Pi> p 2 > Pa) = (Fi (Pi» P2> Pa)> F 2 (p 1? p 2 , p 3 )) 



= (Pi,(P2 2 +p!) 2 ) 



M334 IV.l 

Hence by the chain rule 

dg(v) = g*(v) = f*(F*(v)) = df(F*(v)). 

Now the Jacobian matrix for F is 



13 



F* = 




3F! 9F] 
W 

3F2 
W 




{yZTjf (y 2 +z 2 )* 



Hence at points of M the mapping F* is onto, has rank 2, since either y or z 
is nonzero as M does not intersect the x axis. 

Assume dg is zero at p on M; then dg(v) = for all vectors v tangent to M at 
p. Now if w is tangent vector to E 2 at F(p): then, since F* is onto, 
w = F*(v) for some tangent vector v to E 3 at p. Hence 

df(w) = df(F Jlt (v)) = dg(v) = 0, 
by assumption, and so df is zero at F(p). Now for mappings from E to R 
the analogue of Corollary 1.5.5 gives 

At af A -U 9f ^ 

df = ■*- du +-5- dv. 
ou ov 

Hence 8f/du and 3f/9v are both zero at F(p). But 

F(p) = (Pi,(p 2 2 +P3 2 )*) 
is a point of the profile curve C and so by the definition of an implicitly 
defined curve 9f/3u and df/3v cannot both be zero. This contradiction 
tells us that our original assumption that dg was zero at p is false; dg is never 
zero on M. 



Summary 

Notation 

x 

M 

M : z = f(x, y) 
M : g = c 



Page 124, Definition 1.1 
Page 125, Definition 1.2 
Page 125, line -6 
Page 127, Example 1.3 
Page 127, line -11 



Definitions 

(i) Coordinate patch x : D ►E 3 

(ii) Proper patch x : D ►E 3 

(iii) Surface M 

(iv) Monge patch x(u, v) = (u, v, f(u, v)) 

(v) Simple surface M = x(D) 

(vi) Graph of a function M : z = f(x, y) 

(vii) Implicitly defined surface M : g = c 

(viii) Cylinder 

(ix) Surface of revolution 

(x) Parallel 

(xi) Meridian 



Page 124, 
Page 124, 
Page 125, 
Page 127, 
Page 127, 
Page 127, 
Page 127, 
Page 129, 
Page 129, 
Page 130, 
Page 130, 



Definition 1.1 
line -2 

Definition 1.2 
line 8 
line 11 
Example 1.3 
line -11 
Example 1.5 
Example 1.6 
line 8 
line 9 



hxam 


pies 


(i) 


Unit sphere 2 


(") 


Hemispherical patches on the sphere, 




x(u,v) = (u,v,±(l-u 2 -v 2 )*) 


(iii) 


Sphere 


(iv) 


Circular cylinder, 




M : x 2 + y 2 = r 2 in E 3 



M334 IV. 1 



Page 125, line -3 

Page 126, line 7 
Page 128, line -2 

Page 129, line -13 



Results 
(i) 



(iii) 
(iv) 



Graphs are surfaces. 

An implicitly defined subset M : g = c is a 
surface if the differential dg is never zero 
at any point of M. 

Cylinders are surfaces. 

Surfaces of revolution are surfaces. 



Page 127, Example 1.3 

Page 127, Theorem 1.4 
Page 129, Example 1.5 
Page 129, Example 1.6 



Techniques 

(i) Recognizing a (proper) patch. Page 124, Definition 1.1 (and 

Page 124, line -2) 

(ii) Determining whether subsets of E 3 are surfaces by 

looking for proper patches. Page 125, Definition 1.2 

(iii) Recognizing that a subset of E 3 is a surface since 

it is a graph, a cylinder or a surface of revolution. Page 127, Example 1.3, 

Page 129, Example 1.5, 
Page 129, Example 1.6 

(iv) Determining whether an implicitly defined subset 
of E 3 , M : g = c, is a surface by evaluating the 
differential dg. Page 127, Theorem 1.4 

(v) Sketching surfaces. 



Exercises 

Technique (i) 

1. Page 132, Exercise 4. 



Techniques (ii) and (iii) 

2. Page 131, Exercise 1. Deleting as few points as possible from each subset 

construct a surface. Explain why it is a surface. 



M334 IV.l 
Technique (iv) 

3. Page 132, Exercise 5. 

Technique (v) 

4. Page 132, Exercise 6. 



15 



Solutions 

1. 



Page 132, Exercise 4. 

The Jacobian matrix for the mapping x = (x x , x 2 , x 3 ) : E 2 ►E 3 is 



dxi 


8 Xl \ 


3u~ 


3v ] 


9x 2 


dx 2 


9u 


dv 


3x 3 


9x 3 



(a) 



(b) 



The Jacobian matrix is 

1 °\ 



u 







i 



This always has rank 2 and so x is regular. 
Now if 

x(u, v) = x(u', v') 
then 

(u, uv, v) = (u', u' v', v') 
and so 

u = u', v = v'. 



Hence 


x is one-to-one. 


Since x is one-to-one and regular it is a patch 


The Jacobian matrix is 


J2u 


ol 






3u 2 









o 


ll 





When u = this matrix has rank 1. Hence x is not regular and so is 
not a patch. 



16 



M334IV.1 



(c) 



The Jacob ian matrix is 

h 

2u 

i0 1 + 3v 2 J 



and since v 2 > it has rank 2. Hence x is regular. 

Suppose x(u, v) = x(u', v'). 

Then 

(u,u 2 ,v + v 3 ) = (u',u' 2 ,v' + v' 3 ). 
Hence 

u = u and (v- v') + (v 3 - v' 3 ) = 0. 
It follows that 

(v- v') (l + (v 2 + w' + v' 2 ))*0 
and hence 

(v-v')(l + (v+|) 2 + fv' 2 ) = 0. 

Since 1 + (v + o~) 2 +- 3 f v ' 2 - > it follows that v = v' also and hence x is 
one-to-one. Since x is one-to-one and regular it is a patch, 

(d) If x(u, v) = x(u', v') then 

cos 27ru = cos 27ru', sin 2mi = sin 27ru' and v = v'. 

These conditions are satisfied whenever u = u' + n, for any integer n, 
and v = v'. Hence the mapping is not one-to-one and x is not a patch. 

Page 131, Exercise 1. 

(a) We can write M as M : g = 0, where g(x, y , z) = z 2 - x 2 - y 2 . 

By Corollary 1.5.5 the differential of g is 

dg = -2x dx -2y dy + 2z dz 

and this is zero only at the origin. Hence, by Theorem IV.1.4, the 
cone minus its apex is a surface. 

Since the boundary of a small neighbourhood of the origin is two 
circles we cannot include the origin if we require a surface. 




M334IV.1 17 

(b) Points not on the boundary of the closed disc have the same small 
neighbourhoods as if they were considered as points of the plane 
P : z = 0. Since the differential dz is nonzero it follows from 
Theorem IV. 1.4 that the disc minus its boundary is a surface. 

If we try to map a small open disc continuously to the closed disc in 
such a way that the centre of the open disc is mapped to a point on 
the boundary of the closed disc we find that we have to fold it over. 
Hence the map cannot be one-to-one and so the boundary points 
cannot be included in the surface. 




(c) 




* y 



The region M : xy = 0, x > 0, y > can be written as the union of two 
half planes M t : x = 0, y > and M 2 : y = 0, x > 0. For any point p! 
in the interior of either of these planes it is easy to find a rectangular 
coordinate patch. Thus the subset of M obtained by deleting the z 
axis is a surface. However, for any point p 2 on the z axis, the inter- 
section of these two half planes, it is impossible to find such a patch. 
Any neighbourhood of p 2 has a right angle bend in it and so no 

mapping x : D ►E 3 onto a neighbourhood of p 2 will be differenti- 

able and regular at p 2 . 



18 
3. 



M334IV.1 



Page 132, Exercise 5. 



(a) 



0>) 



IS 



By Theorem IV. 1.4, M : (x 2 + y 2 ) 2 + 3z 2 = 1 is a surface if dg 
never zero on M, where g(x, y, z) - (x 2 + y 2 ) 2 + 3z 2 . Now 

, 9g 3g 3g 

dg = — -dx + — dy + — dz, by Corollary 1.5.5. 
ox dy 9z 

and hence 



dg = 2(x 2 + y 2 ) 2x dx + 2(x 2 + y 2 ) 2y dy + 6z dz 
= 4(x 2 + y 2 ) (x dx + y dy) + 6z dz. 

Hence dg is zero only at the origin and since this is not a point of M 
it follows that M is a surface. 

By Theorem IV. 1.4, M : z(z - 2) + xy = c is a surface if dg is never 
zero on M, where g(x, y, z) = z(z - 2) + xy. 

Now 

dg = 2(z - 1) dz + x dy + y dx 

and hence dg is zero only at the point (0, 0, 1). 

The point (0, 0, 1) belongs to M if and only if 

(z(z- 2) + xy)(0,0, l) = c 
1(1- 2) + 0.0 = c 
c = -l. 

Hence M is a surface except possibly when c = - 1. 

When c = -1, the set M is given by M : z(z - 2) + xy = -1 or, on 
rearranging, M : (z - l) 2 + xy = 0. A sketch of the set M near the 
point (0, 0, 1) shows that it is a 'double' cone. In Solution 2, above, 
we saw that this is not a surface. 




Page 132, Exercise 6. 

M : z = y 3 - 3yx 2 
The intersection of M with the xy plane is given by 

S : = y 3 - 3yx 2 , z = 0. 



M334 IV.l 

Now if 

then 



19 



= y 3 - 3yx 2 

= (y 2 -3x 2 )y 
= (y-V3x)(y+V3x)y. 

The intersection consists of three lines in the xy plane. The lines are y = 0, 
y - y/Sx = and y + V 3x = 0. 

These lines make angles of ^ with each other. 



y = V 3x 



y=V3x 




As a corollary to the intermediate value theorem, we can deduce that 
throughout each of the six resulting regions of the plane the function f(x, y) 
= y 3 - 3yx 2 is always positive or always negative. All we need to do is to 
evaluate f at one point of each region. The values are alternately positive and 
negative. Hence the surface M is alternately above and below the plane. This 
surface is illustrated in Figure 5.22 on page 205. 



20 M334 IV.2 

IV.2 PATCH COMPUTATIONS 



Introduction 

This section follows on directly from Section IV. 1 and assumes familiarity with 
curves, dealt with in Section 1.4, and the dot and cross product, dealt with in 
Section ILL It is also useful to bear in mind the definition of the derivative of a 
mapping, given in Section 1.7. 

Choosing patches enables us to describe surfaces in terms of regions of the plane E 2 . 
In this section we begin to see how we can use the geometry of E 2 to study surfaces. 

This section contains many important examples, which will be used again and again 
throughout the remainder of the course. 

READ; Section IV.2 (pages 133 - 140). 



Comment 



(i) 



Patches 
curves 



Through each point (u ,v ) in the plane E 2 there are two 



ti — >(t,v ) 

and ti — Ku ,t), 

that are parallel to the coordinate axes. 

These are unit-speed curves and their velocities at the point (u , v ) are 
Ui(u ,v ) and U 2 (u , v ). 




ti — ►(t,v ) 



*u 



tl ►(Uq, t) 



The images of these curves under the mapping x are the u- and v- parameter 
curves, v = v and u = u respectively. These parameter curves depend on the 
choice of coordinate patch x. 

Since regular mappings preserve velocity vectors, the partial velocities (the 
velocities of the parameter curves) are given by 



and 



x u( u o» v ) = **{Ui (u , v )) 



*v( u o> v o ) = x *( u 2 (u > v )). 



M334 IV.2 



21 



Hence x u and x v are the composite mappings x 5|e (Ui ) and x J|c (U 2 ). To 
remind ourselves that x u and x v are functions on E 2 we sometimes write 
them as x u (u, v) and x v (u, v). To calculate these explictly we look at the 
parameter curves. The u- parameter curve v = v is given by 



a : ti — ►x(t, v ) = (x! (t, v ), x 2 (t, v ), x 3 (t, v )). 



Hence 



da 
x u (u ,v ) = ^( u o) atx(u ,v ) 



/9xi , v 3x 2 . . dx 3 

=( r— (u , v ), — (u , v ), — (u , v )) at x(u , v ). 
ou ou ou 



Since x is regular, the derivate x* is one-to-one and hence it carries linearly 
independent vectors into linearly independent vectors. The tangent vectors 
Ui (u , v ) and U 2 (u , v ) are linearly independent in E 2 and so x u (u , v ) 
and x v (u , v ) are linearly independent in E 3 . In Section II. 1 we saw that 
this is equivalent to the condition 

x u( u o> v o) x x v (u , v ) # 0. 

That is, x u X x v is never zero. 

In fact this condition that x u X x v is never zero is enough to ensure that x is 
regular. The condition ensures that the images of U t (u , v ) and U 2 (u , v ) 
under x* are linearly independent and this implies that x* is one-to-one at 
each point. 

When O'Neill comes to calculate x u X x v he obtains the equation 



X^j X Xy ~ 



Ui 

9xj 
3u~ 

3xj 
3v~ 



U 2 

3x 2 
3u~ 

9x 2 
9v~ 



U 3 

8x3 
3u 

8x3 
bv 



Here L^ , U 2 , U3 are vector valued functions with domain E 3 while all the 
other functions have domain E 2 and hence the equation is not technically 
correct. However it is very convenient to use such expressions as long as you 
are always aware of how they could be made rigorous. To make it rigorous 
we could replace Ui, U 2 , U 3 by the composite functions Uj o x, U 2 ° x and 
U 3 © x, or state that if the functions with the domain E 2 are evaluated at the 
point (u , v ) then those with domain E 3 are evaluated at the point 
p = x(u ,v ). 



22 

Supplementary Comments 

(i) Page 136: line- 1 



M334 IV.2 



(") 



r X|j X Xy — 



U t 



- cos v sin u 



U, 



COS V COS u 



sin v cos u - sin v sin u 



U 3 



cos v 



= cos 2 v cos u Ui - cos v (-cos v sin u) U 2 
+((-cosv sinu) (-sin v sinu) - (-sin v cos u)(cos v cos u)) U 3 

«= cos 2 v cosu Uj * cos 2 v sinu U 2 + cos v sin v U 3 
since cos 2 u + sin 2 u = 1. 

Page 139: line 2, x u and x v are linearly independent 
Here 

x u( u » v ) = (g'( u )> h '( u ) cos v > h '( u ) sin v ) 
x v (u, v) = (0, -h(u) sin v, h(u) cos v) 

and hence 

x u (u, v) X x v (u, v) = (h(u) h'(u), -h(u)g'(u) cos v, -h(u)g'(u) sin v). 
If x u (u, v) and x v (u, v) are linearly dependent at x(u , v ) then 

x u (u , v ) X x v (u , v ) = 0. 

But since h(u ) > and cos v and sin v cannot simultaneously be zero, we 
would have h'(u ) = g'(u ) = 0. This is impossible since by the definition (on 
page 21) a parametrization is a regular curve. Hence x u (u, v) X x v (u, v) is 
never zero and the partial velocities x u (u, v) and x v (u, v) are linearly 
independent. 



Additional Text 

There are two important classes of surfaces defined in the Exercises at the end of 
Section IV.2. 



(i) Ruled Surfaces 

READ the definition on page 140 between Exercises 3 and 4 and look at the 
examples in Exercises 5 and 6. 

Comment |3(u) gives the position along the base curve, 6(u) gives the 'direction' of 
the line L and v gives the 'distance' along this line from the base curve. 



M334 IV.2 



23 



x(u, v) 




origin 



5(u) need not be the same length and direction for each u. 



(ii) Quadric surfaces 

READ the definition on page 142, between Exercises 9 and 10, and browse 
through the examples in Exercises 10 to 11. 



Summary 



Notation 



x u (u ,v ) and x v (u ,v ) 



x u and x v 



Page 134, Definition 2.1 
Page 134, line 12 



Definitions 

(i) u- and v- parameter curves, v = v and u = u 

(ii) Partial velocities, x u (u ,v ) and x v (u ,v ) 

(iii) Parametrization 

(iv) Cross-sectional curve of a cylinder 

(v) Rulings of a cylinder 

(vi) Ruled surface 

(vii) Rulings 

(viii) Ruled form 

(ix) Base curve 

(x) Director curve 

(xi) Quadric surface 



Page 133, line -9 
Page 134, Definition 2.1 
Page 135, Definition 2.3 
Page 138, line 9 
Page 138, line 11 



► Page 140, lines - 16 to - 8 



Page 142, line 16 



24 M334 IV.2 

Examples 

(i) The geographical patch on the sphere: 

x(u, v) = (r cos v cos u, r cos v sin u, r sin v), 
where 

uG (-7T, 7r), vG (-£, «) Page 135, line 3 



(ii) Parametrization of a cylinder: 

x(u, v) = («! (u), ol 2 (u), v) Page 137, Example 2.4 

(iii) Parametrization of a surface of revolution: 

x(u, v) = (g(u), h(u) cos v, h(u) sin v) Page 138, Example 2.5 

(iv) Parametrization of a torus of revolution: 

x(u, v) = ((R + r cos u) cos v, 

(R + r cos u) sin v, r sin u) Page 139, Example 2.6 

(v) Parametrization of a ruled surface in ruled form: 

x(u, v) = j8(u) + v5(u) Page 140, line - 12 

(vi) Helicoid: 

x(u, v) = (u cos v, u sin v, bv) Page 141, Exercise 7 



Results 

(i) x : D ►E 3 is regular if and only if x u X x v is 

never zero. Page 136, line 6 

(ii) The partial velocities are the images of the standard 
basis vectors under the derivative mapping x^: 

x*(Ui ) = x u , x„ c (U 2 ) = x v . Text, page 20 



Techniques 

(i) Description of the parameter curves. Page 133, line -9 

(ii) Evaluation of partial velocities. Page 134, line -7 

(iii) Testing regularity of a parametrization. Page 136, line 6 

(iv) Use of standard parametrization of cylinders and 

surfaces of revolution. Page 137, Example 2.4, 

Page 138, Example 2.5 
(v) Finding the ruled form of a surface, in certain 

simple examples. Page 140, line - 16 

(vi) Familiarity with the shapes and equations of 

quadric surfaces. Page 142, line 16 



M334 IV.2 
Exercises 



25 



Techniques (i), (ii) and (Hi) 

1. Page 141, Exercise 7. 

Technique (Hi) 

2. Page 140, Exercise 2. 

Technique (iv) 

3. Page 141, Exercise 6. 

4. Page 140, Exercise 1. 

Technique (v) 

5. Page 140, Exercise 4. 

Technique (vi) 

6. Page 142, Exercises 10 and 11. Only sketch the surfaces. 

7. Page 142, Exercise 12(a) and (b). 



Solutions 



1. Page 141, Exercise 7. 



(a) 



(b) 



To check that x is a patch we must show that it is one-to-one and 
regular. If x(u, v) = x(u', v') then 

u cos v = u' cos v', u sin v = u' sin v' and v = v\ 

Since v = v' and the cosine and sine functions cannot be simul- 
taneously zero it follows that u = u'. Hence x is one-to-one. 

Now 



Xy X Xy — 



U, 



cos V 



-u sin v 



U, 



sin v 



u cos v 



U 3 



b 



= (b sin v, - b cos v, u). 

Hence ||x u X x v || 2 = b 2 + u 2 > 0, and so x u X x v i= 0. Thus x is 
regular and therefore a patch. 

The u- parameter curve v = v is 

1 1 — >x(t, v ) = (t cos v , t sin v , bv ) 

= (0, 0, bv ) + t(cos v , sin v , 0), 

which is a straight line parallel to the line in the xy plane making an 
angle v with the x axis. 



26 



M334 IV.2 




The v-parameter curve u = u is 

ti — »"x(u , t) = (u cos t, u sin t, bt). 

As t increases by 2flr the curve goes once round the z axis, in a circle 
of radius u , and rises 27rb. The v-parameter curves are helices about 
the z axis. 




(c) In order to express the helicoid in implicit form we must find some 

function g and constant c such that g(p) = c if and only if p = 
x(u , v ) for some (u , v ). 

If p = x(u ,v )then 



and 



p 3 =bv , v =p 3 /b 



Pi =u cos(p 3 /b), p 2 =u sin(p 3 /b). 



M334 IV.2 27 

Hence for each point p of the helicoid 
Pi sin (p 3 /b) = p 2 cos (p 3 /b) 

so if a point belongs to the helicoid it also belongs to the implicitly 
defined surface 

M : x sin(z/b) - y cos(z/b) = 0. 

We need to show conversely that any point of M belongs to the 
helicoid. Suppose p satisfies the above relationship. Then since b # 
we can set v = p 3 /b and we obtain 

p! sin v =P2 cos v . 

For any value of v at least one of cos v , sin v must be nonzero. 
Suppose cosvq^O. Then we can find u =pi/cosv . Now 
p 2 cos v = Pi sin v = u cos v sin v and so, since cos v =£ 0, 
p 2 =u sin v . Hence we can write p in the form (u cos v , u sin v , 
bv ) and so it belongs to the helicoid. 

A similar argument works when sin v =£ 0. 
Hence the helicoid is the surface 

M : x sin(z/b) - y cos(z/b) = 0. 

2. Page 140, Exercise 2. 

By Lemma II. 1.8, 

|| x u X x v || 2 = (x u • x u ) (x v • x v ) - (x u • x v ) 2 
*=EG - F 2 . 

Now x is regular if and only if x u X x v =£ 0. But x u X x v is never zero if and 
only if ||x u X Xyfl is never zero, which is true if and only if EG - F 2 is never 
zero. 



3. Page 141, Exercise 6. 

x(u, v) = 0(u) + vq. 
Hence x u = |3' 
and x v = q: 

x u X u v = 0' X q, 

and so x is regular if and only if |3' X q is never zero. 

In Example 2.4 the curve is in the xy plane and q = (0, 0, 1). 



Page 140, Exercise 1. 

(a) A parametrization of the profile curve is ui — ►(u, cosh u, 0). 
A parametrization of the catenoid is 

x : (u, v)i — ►(u, cosh u cos v, cosh u sin v) 

since rotation about the x axis keeps the x coordinate fixed. 



28 



M334 IV.2 



u, cosh u, 0) 




(b) 



x(u, v) = (u, cosh u cos v, cosh u sin v) 



C : (z - 2) 2 + y 2 = 1 is a circle in the yz plane of radius one and 
centre (0, 2). A possible parametrization is 

ui — ►(O, sin u, 2 + cos u). 
A parametrization of the torus is 

x : (u, v)i — >((2 + cos u) sin v, sin u, (2 + cos u) cos v), 
since rotation about the y axis keeps the y coordinate fixed. 

y 




((2 + cos u) sin v, sin u, (2 + cos u) cos v) 



, sin u, 2 + cos u) 



>z 



(c) We cannot follow the same procedure as in parts (a) and (b) since the 

profile curve cuts the axis of rotation and hence the standard 
'parametrization' will not be regular. However we can describe the 
surface in terms of a Monge patch. 



M334 IV.2 



29 



P = (Pl»P2»P3) 




A point is on the paraboloid M if the square of its distance from the 
z axis is equal to its height above the xy plane. Thus p = (p x , p 2 , P3 ) 
belongs to M if and only if pj + p\ = p 3 . Hence M : z = x 2 + y 2 and 
we can parametrize it by mapping each point in the xy plane to the 
unique point on the surface above it. This gives the single patch 

x : (u, v)i — >>(u, v, u 2 + v 2 ). 



Page 140, Exercise 4. 

Look at the intersection of M with the plane x = c, for some constant c. This 
is the straight line z = cy in the plane x = c. It passes through the points 
(c, 0, 0) and (c, 1, c). We can write it as 

v.— *(c,0,0) + v(0, l,c). 

Letting the point (c, 0, 0) vary along the x axis we obtain the para- 
metrization 



(u,v)i — >(u, 0, 0) + v(0, l,u), 



which is in ruled form. 

Similarly looking at the intersection of M with planes of the form y = c we 
find the parametrization 

(u,v)i— ►(0,v,0)+u(l > 0,v). 

Alternatively, since M : z = xy, M is described by the Monge patch 

x : (u, v)i ►(u, v, uv), 

from which the above two ruled forms follow. 



30 
6. Page 142, Exercises 10 and 11. 

10(a). Ellipsoid 



M334 IV.2 



A z 




10(b). Elliptic hyperboloid 




M334 IV.2 

10(c). Elliptic hyperboloid (two sheets) 



31 




1 1 . Elliptic parab oloid 




32 

7. 



M334 IV.2 



Page 142, Exercise 12 
(a) 



(b) 



The mapping x : E 2 ► E 3 does map into M since 

_ (a(u + v)) 2 (b(u-v)) 2 
a 2 b 2 * 

We need to show further that it maps onto M. For any point p 
belonging to M we need to solve the equations 

Pi = a(u + v ) 
p 2 =b(u - v ) 
p 3 =4u v . 
We can solve the first two, obtaining 
_Pi , P2 _Pl P2 

Uo = 2a" 2b' Vo= 2a~2b 
and since p belongs to M the third equation is also satisfied. 
Next we need to check that x is one-to-one. Suppose 

x(u, v) =x(u', v'): 
then 

U + V = U + V 
I I 

u - v = u - V 
and hence u = u', v = v\ The mapping x is one-to-one. 
Now we need to check that x is regular: we use the usual method. 



x u x x u ~ 



Ui 



u 2 

b 



U 3 

4v 



a -b 4u 

= (4b(u + v), 4a(v - u), - 2ab). 



Now, for the definition of the hyperbolic paraboloid to be meaning- 
ful we need a,b > 0. Hence -2ab ¥= and so x u X x v is never zero. 
This in turn proves that x u and x v are linearly independent and 
hence x is regular. 

Thus x is indeed a patch and to complete the proof we need only 
prove that x is proper. To prove that x is proper we have to show 
that the inverse function x - l 
given by the formula 



Mi ►E 2 is continuous. Since x 1 is 



JL + JL 

2a 2b 



x 

2a" 



_y_ 

2b 



continuity follows from standard results in analysis. 

We can write x in ruled form as follows 

(i) x(u, v) = (au, bu, 0)+v(a,-b,4u) 

and 

(ii) x (u, v) = (av, -bv, 0) + u(a, b, 4v). 



M334 IV.3 33 

IV.3 DIFFERENTIABLE FUNCTIONS AND TANGENT VECTORS 

Introduction 



This section follows on from the previous section, IV.2, and follows the develop- 
ment of Chapter I, Sections 1 to 4. You will also need the chain rule dealt with in 
Section 1.7. 

The section begins with certain rather technical but intuitively obvious definitions 
and results. Mappings that you would expect to be differentiable turn out to be 
differentiable, and, given a coordinate patch, we can use it to give explicit descrip- 
tions of curves in a surface. Do not worry about the technicalities. 

We then show that the partial velocities, for a given patch, do tell us something 
independent of the choice of patch. The tangents to any curve in the surface can be 
written as a linear combination of them. These tangents form the tangent planes to 
the surface. We use them to define a directional derivative for functions defined only 
on the surface, and not on the whole of E 3 . 



READ: Section IV.3 (pages 143-149). Do not spend much time on pages 
143 - 145. Just try to understand the results. 



Comments 

(i) Differentiation on M and on E 3 We have now met several different defini- 

tions of differentiation with domain or codomain either E 3 or some surface 

M. If we are dealing with a function F : E n ►E 111 between Euclidean spaces 

and we are using the definition of differentiability given in Section 1.7 we 
say that F is Euclidean differentiable. If we are using either of the definitions 
of this section for functions with domain or codomain some surface M we 

indicate it by saying that G : M >E n is differentiable or F : E n »M is 

differentiable. We need to compare these definitions of differentiability with 
the Euclidean definition. 

Directly from the definitions we have that G : M ►E 11 or F : E n ►M are 

differentiable if their respective coordinate expressions are Euclidean 
differentiable. 

Suppose we have a function F : E n ►M. Then we can also consider it as a 

function F : E n ►E 3 . If we know that F is differentiable as a function 

from E n to M then we would hope that F is also Euclidean differentiable. It 
is. Since F : E n - — >M is differentiable it follows, by definition, that 

x _1 o F : E n ►E 2 is Euclidean differentiable, for each patch x. But by